Sequences & series

1. Definitions & types


ID is: 2794 Seed is: 7933

Completing a sequence using the general formula

The general term is given for each sequence below. Calculate the missing term(s).

  1. 11;21;31;c;51;

    Tn=10n1

    Answer:

    c=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What is the position of the term that you want to find? The position is the value of 'n' that we substitute into the general formula.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the missing term, we use the equation for the general term:

    Tn=10n1

    For the missing term c, we have n=4:

    T4=−10(4)1c=41

    Submit your answer as:
  2. 10;q;22;y;34;

    Tn=6n4

    Answer:
    1. q=
    2. y=
    numeric
    numeric
    2 attempts remaining
    STEP: <no title>
    [−4 points ⇒ 0 / 4 points left]

    To find the two missing terms, we use the equation for the general term:

    Tn=6n4

    For the missing term q, we have n=2:

    T2=6(2)4q=16

    For the missing term y, we have n=4:

    T4=6(4)4y=28

    Submit your answer as: and

ID is: 2794 Seed is: 1730

Completing a sequence using the general formula

The general term is given for each sequence below. Calculate the missing term(s).

  1. 0;2;4;b;8;

    Tn=2n+2

    Answer:

    b=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What is the position of the term that you want to find? The position is the value of 'n' that we substitute into the general formula.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the missing term, we use the equation for the general term:

    Tn=2n+2

    For the missing term b, we have n=4:

    T4=−2(4)+2b=6

    Submit your answer as:
  2. 5;q;29;z;53;

    Tn=12n+7

    Answer:
    1. q=
    2. z=
    numeric
    numeric
    2 attempts remaining
    STEP: <no title>
    [−4 points ⇒ 0 / 4 points left]

    To find the two missing terms, we use the equation for the general term:

    Tn=12n+7

    For the missing term q, we have n=2:

    T2=12(2)+7q=17

    For the missing term z, we have n=4:

    T4=12(4)+7z=41

    Submit your answer as: and

ID is: 2794 Seed is: 1458

Completing a sequence using the general formula

The general term is given for each sequence below. Calculate the missing term(s).

  1. 4;2;0;b;4;

    Tn=2n+6

    Answer:

    b=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What is the position of the term that you want to find? The position is the value of 'n' that we substitute into the general formula.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the missing term, we use the equation for the general term:

    Tn=2n+6

    For the missing term b, we have n=4:

    T4=−2(4)+6b=2

    Submit your answer as:
  2. 9;t;33;y;57;

    Tn=12n3

    Answer:
    1. t=
    2. y=
    numeric
    numeric
    2 attempts remaining
    STEP: <no title>
    [−4 points ⇒ 0 / 4 points left]

    To find the two missing terms, we use the equation for the general term:

    Tn=12n3

    For the missing term t, we have n=2:

    T2=12(2)3t=21

    For the missing term y, we have n=4:

    T4=12(4)3y=45

    Submit your answer as: and

ID is: 2782 Seed is: 3172

Identifying an arithmetic sequence

Given the list of numbers: 5;12;19;26;33;

Calculate the common difference (if there is one).

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The common difference is the difference between any two successive terms of an arithmetic sequence. If a sequence does not have a common difference between successive terms, then it is not arithmetic.
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The common difference (d) is calculated by finding the difference between any two successive terms:

d=T2T1ord=T3T2ord=T4T3

Important: d is NOT equal to T1T2 or T2T3.

Term T1 T2 T3 T4 T5
Value of term 5 12 19 26 33


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We calculate the common difference (d) at least two different ways (to show that it is constant throughout the sequence):

d=T2T1=12(5)=7d=T3T2=19(12)=7d=T4T3=26(19)=7

All the values of d are equal, which means we have found the common difference and therefore the sequence is arithmetic.


Submit your answer as:

ID is: 2782 Seed is: 3398

Identifying an arithmetic sequence

Given the list of numbers: 8;9;15;17;21;

Determine the common difference (if there is one).

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The common difference is the difference between any two successive terms of an arithmetic sequence. If a sequence does not have a common difference between successive terms, then it is not arithmetic.
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The common difference (d) is calculated by finding the difference between any two successive terms:

d=T2T1ord=T3T2ord=T4T3

Important: d is NOT equal to T1T2 or T2T3.

Term T1 T2 T3 T4 T5
Value of term 8 9 15 17 21


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We calculate the common difference (d) at least two different ways (to show that it is constant throughout the sequence):

d=T2T1=9(8)=1d=T3T2=15(9)=6d=T4T3=17(15)=2

We see that the results are not the same - the difference between terms is not 'common.' Therefore, the sequence is not arithmetic.


Submit your answer as:

ID is: 2782 Seed is: 2241

Identifying an arithmetic sequence

Given the following list of numbers: 8;11;17;21;26;

Find the common difference (if there is one).

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The common difference is the difference between any two successive terms of an arithmetic sequence. If a sequence does not have a common difference between successive terms, then it is not arithmetic.
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The common difference (d) is calculated by finding the difference between any two successive terms:

d=T2T1ord=T3T2ord=T4T3

Important: d is NOT equal to T1T2 or T2T3.

Term T1 T2 T3 T4 T5
Value of term 8 11 17 21 26


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We calculate the common difference (d) at least two different ways (to show that it is constant throughout the sequence):

d=T2T1=11(8)=3d=T3T2=17(11)=6d=T4T3=21(17)=4

We see that the results are not the same - the difference between terms is not 'common.' Therefore, the sequence is not arithmetic.


Submit your answer as:

ID is: 2675 Seed is: 5402

Completing the sequence

Write down the next three terms in each of the following arithmetic sequences:

  1. 35;15;5;

    Answer:35;15;5; ; ;
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the common difference:

    d=T2T1 or d=T3T2

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=15(35)=20 or d=T3T2=5(15)=20Therefore, T4=25T5=45T6=65

    Submit your answer as: andand
  2. 9.5;15.5;21.5;

    Answer:9.5;15.5;21.5; ; ;
    numeric
    numeric
    numeric
    2 attempts remaining
    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=15.5(9.5)=6 or d=T3T2=21.5(15.5)=6Therefore, T4=27.5T5=33.5T6=39.5

    Submit your answer as: andand
  3. 4v;5v;6v;

    Answer:4v;5v;6v; ; ;
    polynomial
    polynomial
    polynomial
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    This question involves variables, but the same method is used to find the common difference and to complete the sequence as the previous two examples.
    STEP: <no title>
    [−6 points ⇒ 0 / 6 points left]

    To complete the sequence, we need to determine the common difference in terms of v:

    d=T2T1=5v(4v)=v or d=T3T2=6v(5v)=vTherefore, T4=7vT5=8vT6=9v

    Submit your answer as: andand

ID is: 2675 Seed is: 5831

Completing the sequence

Write down the next three terms in each of the following arithmetic sequences:

  1. 29;49;69;

    Answer:29;49;69; ; ;
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the common difference:

    d=T2T1 or d=T3T2

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=49(29)=20 or d=T3T2=69(49)=20Therefore, T4=89T5=109T6=129

    Submit your answer as: andand
  2. 19.7;35.7;51.7;

    Answer:19.7;35.7;51.7; ; ;
    numeric
    numeric
    numeric
    2 attempts remaining
    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=35.7(19.7)=16 or d=T3T2=51.7(35.7)=16Therefore, T4=67.7T5=83.7T6=99.7

    Submit your answer as: andand
  3. 11v;29v;47v;

    Answer:11v;29v;47v; ; ;
    polynomial
    polynomial
    polynomial
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    This question involves variables, but the same method is used to find the common difference and to complete the sequence as the previous two examples.
    STEP: <no title>
    [−6 points ⇒ 0 / 6 points left]

    To complete the sequence, we need to determine the common difference in terms of v:

    d=T2T1=29v(11v)=18v or d=T3T2=47v(29v)=18vTherefore, T4=65vT5=83vT6=101v

    Submit your answer as: andand

ID is: 2675 Seed is: 8117

Completing the sequence

Write down the next three terms in each of the following arithmetic sequences:

  1. 21;12;3;

    Answer:21;12;3; ; ;
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the common difference:

    d=T2T1 or d=T3T2

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=12(21)=9 or d=T3T2=3(12)=9Therefore, T4=6T5=15T6=24

    Submit your answer as: andand
  2. 4.3;15.7;35.7;

    Answer:4.3;15.7;35.7; ; ;
    numeric
    numeric
    numeric
    2 attempts remaining
    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=15.7(4.3)=20 or d=T3T2=35.7(15.7)=20Therefore, T4=55.7T5=75.7T6=95.7

    Submit your answer as: andand
  3. 18c;15c;12c;

    Answer:18c;15c;12c; ; ;
    polynomial
    polynomial
    polynomial
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    This question involves variables, but the same method is used to find the common difference and to complete the sequence as the previous two examples.
    STEP: <no title>
    [−6 points ⇒ 0 / 6 points left]

    To complete the sequence, we need to determine the common difference in terms of c:

    d=T2T1=15c(18c)=3c or d=T3T2=12c(15c)=3cTherefore, T4=9cT5=6cT6=3c

    Submit your answer as: andand

ID is: 3089 Seed is: 3100

Arithmetic sequences with algebraic terms

Given the following arithmetic sequence:

3b5m;5b5m;7b5m;9b5m;
  1. Calculate the common difference (d).

    Answer: d=
    expression
    2 attempts remaining
    STEP: Calculate the common difference (d).
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 4 terms of a sequence and specifically tells us that the terms follow an arithmetic pattern. We must find the common difference of the sequence, d.

    This is an example of a non-numeric arithmetic sequence, where each term in the given sequence is an algebraic expression. To find the common difference (d), we subtract any two successive terms in the sequence:

    d=T2T1=(5b5m)(3b5m)=2b

    We could also have chosen the following terms:

    d=T3T2=(7b5m)(5b5m)=2b

    As expected, both calculations give the same value for d.

    Therefore, the common difference between each term in the sequence is 2b.


    Submit your answer as:
  2. Determine the next 2 terms in the sequence.

    Answer:

    T5=

    T6=

    expression
    expression
    2 attempts remaining
    STEP: Determine the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    We have the common difference of the sequence and the first 4 terms, and the question asks us to find the next 2 terms.

    To determine any term in the sequence, we add the common difference to the term before it. In other words: Tn=Tn1+d. Thus to find the fifth term, add the common difference to the fourth term.

    T5=T4+2b=(9b5m)+2b=11b5m

    Repeat the calculation to continue to the next term:

    T6=T5+2b=(11b5m)+2b=13b5m

    Therefore, the required terms are: T5=11b5m and T6=13b5m.


    Submit your answer as: and

ID is: 3089 Seed is: 9953

Arithmetic sequences with algebraic terms

For the following arithmetic sequence:

3h+9u;3h+7u;3h+5u;3h+3u;
  1. Find d, the common difference of the sequence.

    Answer: d=
    expression
    2 attempts remaining
    STEP: Find d, the common difference of the sequence.
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 4 terms of a sequence and specifically tells us that the terms follow an arithmetic pattern. We must find the common difference of the sequence, d.

    This is an example of a non-numeric arithmetic sequence, where each term in the given sequence is an algebraic expression. To find the common difference (d), we subtract any two successive terms in the sequence:

    d=T2T1=(3h+7u)(3h+9u)=2u

    We could also have chosen the following terms:

    d=T3T2=(3h+5u)(3h+7u)=2u

    As expected, both calculations give the same value for d.

    Therefore, the common difference between each term in the sequence is 2u.


    Submit your answer as:
  2. Find the next 2 terms in the sequence.

    Answer:

    T5=

    T6=

    expression
    expression
    2 attempts remaining
    STEP: Find the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    We have the common difference of the sequence and the first 4 terms, and the question asks us to find the next 2 terms.

    To determine any term in the sequence, we add the common difference to the term before it. In other words: Tn=Tn1+d. Thus to find the fifth term, add the common difference to the fourth term.

    T5=T42u=(3h+3u)2u=3h+u

    Repeat the calculation to continue to the next term:

    T6=T52u=(3h+u)2u=3hu

    Therefore, the required terms are: T5=3h+u and T6=3hu.


    Submit your answer as: and

ID is: 3089 Seed is: 2932

Arithmetic sequences with algebraic terms

The following arithmetic sequence is given:

10q+s;13q+s;16q+s;
  1. Calculate the common difference (d).

    Answer: d=
    expression
    2 attempts remaining
    STEP: Calculate the common difference (d).
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 3 terms of a sequence and specifically tells us that the terms follow an arithmetic pattern. We must find the common difference of the sequence, d.

    This is an example of a non-numeric arithmetic sequence, where each term in the given sequence is an algebraic expression. To find the common difference (d), we subtract any two successive terms in the sequence:

    d=T2T1=(13q+s)(10q+s)=3q

    We could also have chosen the following terms:

    d=T3T2=(16q+s)(13q+s)=3q

    As expected, both calculations give the same value for d.

    Therefore, the common difference between each term in the sequence is 3q.


    Submit your answer as:
  2. Calculate the next 2 terms in the sequence.

    Answer:

    T4=

    T5=

    expression
    expression
    2 attempts remaining
    STEP: Calculate the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    We have the common difference of the sequence and the first 3 terms, and the question asks us to find the next 2 terms.

    To determine any term in the sequence, we add the common difference to the term before it. In other words: Tn=Tn1+d. Thus to find the fourth term, add the common difference to the third term.

    T4=T33q=(16q+s)3q=19q+s

    Repeat the calculation to continue to the next term:

    T5=T43q=(19q+s)3q=22q+s

    Therefore, the required terms are: T4=19q+s and T5=22q+s.


    Submit your answer as: and

ID is: 2967 Seed is: 535

Arithmetic sequences: finding the common difference

Consider the following sequence:

15;8;1;6
  1. Find the common difference.
    Answer: d=
    numeric
    2 attempts remaining
    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    To find the common difference, we subtract any two successive terms in the sequence. Let us use the third term (T3) and the fourth term (T4). (We could also have chosen the second term (T2) and the third term (T3), both should give us the same answer.)

    We find the difference between any term in the sequence and the term before it:

    d=T4T3=6(1)=6+1=7

    We can check that d is constant by subtracting two other terms in the sequence:

    d=T3T2=1(8)=1+8=7

    Therefore, the common difference is 7.


    Submit your answer as:
  2. Write down the next three terms of the sequence.
    Answer:

    T5=
    T6=
    T7=

    numeric
    numeric
    numeric
    2 attempts remaining
    STEP: Calculate the next term in the sequence
    [−1 point ⇒ 2 / 3 points left]

    Add the common difference to the last known term to get the next term in the sequence:

    T5=T4+d=6+7=13


    STEP: Calculate the following two terms
    [−2 points ⇒ 0 / 3 points left]

    Using the same method we can calculate the following two terms in the sequence:

    T6=T5+d=13+7=20T7=T6+d=20+7=27

    Therefore the sequence is,

    15;8;1;6;13;20;27

    Submit your answer as: andand

ID is: 2967 Seed is: 1317

Arithmetic sequences: finding the common difference

  1. Find the common difference.
    5;3;1
    Answer: d=
    numeric
    2 attempts remaining
    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    To find the common difference, we subtract any two successive terms in the sequence. Let us use the second term (T2) and the third term (T3). (We could also have chosen the first term (T1) and the second term (T2), both should give us the same answer.)

    We find the difference between any term in the sequence and the term before it:

    d=T3T2=13=2

    We can check that d is constant by subtracting two other terms in the sequence:

    d=T2T1=35=2

    Therefore, the common difference is 2.


    Submit your answer as:
  2. Calculate the next four terms of the sequence.
    Answer:

    T4=
    T5=
    T6=
    T7=

    numeric
    numeric
    numeric
    numeric
    2 attempts remaining
    STEP: Calculate the next term in the sequence
    [−1 point ⇒ 3 / 4 points left]

    Add the common difference to the last known term to get the next term in the sequence:

    T4=T3+d=12=1


    STEP: Calculate the following three terms
    [−3 points ⇒ 0 / 4 points left]

    Using the same method we can calculate the following three terms in the sequence:

    T5=T4+d=12=3T6=T5+d=32=5T7=T6+d=52=7

    Therefore the sequence is,

    5;3;1;1;3;5;7

    Submit your answer as: andandand

ID is: 2967 Seed is: 7103

Arithmetic sequences: finding the common difference

Consider the following sequence:

10;23;36;49
  1. Calculate the common difference (d).
    Answer: d=
    numeric
    2 attempts remaining
    STEP: Calculate the common difference (d).
    [−1 point ⇒ 0 / 1 points left]

    To find the common difference, we subtract any two successive terms in the sequence. Let us use the first term (T1) and the second term (T2). (We could also have chosen the second term (T2) and the third term (T3), both should give us the same answer.)

    We find the difference between any term in the sequence and the term before it:

    d=T2T1=2310=13

    We can check that d is constant by subtracting two other terms in the sequence:

    d=T3T2=3623=13

    Therefore, the common difference is 13.


    Submit your answer as:
  2. Determine the next four terms of the sequence.
    Answer:

    T5=
    T6=
    T7=
    T8=

    numeric
    numeric
    numeric
    numeric
    2 attempts remaining
    STEP: Calculate the next term in the sequence
    [−1 point ⇒ 3 / 4 points left]

    Add the common difference to the last known term to get the next term in the sequence:

    T5=T4+d=49+13=62


    STEP: Calculate the following three terms
    [−3 points ⇒ 0 / 4 points left]

    Using the same method we can calculate the following three terms in the sequence:

    T6=T5+d=62+13=75T7=T6+d=75+13=88T8=T7+d=88+13=101

    Therefore the sequence is,

    10;23;36;49;62;75;88;101

    Submit your answer as: andandand

ID is: 2741 Seed is: 9337

Working with the common difference

  1. Compute the value of z for the following arithmetic sequence:

    2z31;8z3+1;11z3;

    If the answer is a non-integer, write the answer as a simplified fraction.

    Answer: z=
    fraction
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    The key to starting this question is the fact that the terms form an arithmetic sequence: that means that the difference between the first and second terms is equal to the difference between the second and third terms.


    STEP: Set up an equation based on the common difference of the terms
    [−1 point ⇒ 3 / 4 points left]

    The given sequence is arithmetic, so the difference between successive terms will be the same (common). So we know that T2T1=T3T2 and we can solve for z:

    T2T1=T3T2(8z3+1)(2z31)=(11z3)(8z3+1)

    STEP: Simplify the equation
    [−2 points ⇒ 1 / 4 points left]

    For an equation with fractions, we can multiply both sides of the equation by the LCD of the fractions to remove the denominators. In this case, the LCD is 3.

    3(8z3+1)3(2z31)=3(11z3)3(8z3+1)8z+3(2z3)=11z(8z+3)6z+6=3z3

    STEP: Solve for z
    [−1 point ⇒ 0 / 4 points left]

    Solve for z:

    3z=9 z=3

    Therefore, z=3.


    Submit your answer as:
  2. Determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions, not decimals.

    Answer: ; ; ;
    fraction
    fraction
    fraction
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The first three terms in the sequence are given in terms of z. We have calculated the value of z, so we can substitute and simplify to find the terms.


    STEP: Substitute z=3 into the expressions for the terms
    [−3 points ⇒ 0 / 3 points left]

    To find the values of the terms, we need to substitute z=3 into the expressions for each of the three terms:

    \[ \begin{aligned} \text{First term: } \quad T_{1} & = \frac{2 z}{3} - 1 \\ & = \frac{2 \left( -3 \right)}{3} - 1 \\ & = -3 \\ \\ \text{Second term: } \quad T_{2} & = \frac{8 z}{3} + 1 \\ & = \frac{8 \left( -3 \right)}{3} + 1 \\ & = -7 \\ \\ \text{Third term: } \quad T_{3} & = \frac{11 z}{3} \\ & = \frac{11 \left( -3 \right) \\ & = -11 \end{aligned} \]

    Therefore, the first three terms of the sequence are: 3,7 and 11.


    Submit your answer as: andand

ID is: 2741 Seed is: 9510

Working with the common difference

  1. Calculate y for the following arithmetic sequence:

    3y21;3y22;11y26;

    If the answer is a non-integer, write the answer as a simplified fraction.

    Answer: y=
    fraction
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    The key to starting this question is the fact that the terms form an arithmetic sequence: that means that the difference between the first and second terms is equal to the difference between the second and third terms.


    STEP: Set up an equation based on the common difference of the terms
    [−1 point ⇒ 3 / 4 points left]

    The given sequence is arithmetic, so the difference between successive terms will be the same (common). So we know that T2T1=T3T2 and we can solve for y:

    T2T1=T3T2(3y22)(3y21)=(11y26)(3y22)

    STEP: Simplify the equation
    [−2 points ⇒ 1 / 4 points left]

    For an equation with fractions, we can multiply both sides of the equation by the LCD of the fractions to remove the denominators. In this case, the LCD is 2.

    2(3y22)2(3y21)=2(11y26)2(3y22)3y4(3y2)=11y12(3y4)6y2=8y8

    STEP: Solve for y
    [−1 point ⇒ 0 / 4 points left]

    Solve for y:

    2y=6 y=3

    Therefore, y=3.


    Submit your answer as:
  2. Determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions, not decimals.

    Answer: ; ; ;
    fraction
    fraction
    fraction
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The first three terms in the sequence are given in terms of y. We have calculated the value of y, so we can substitute and simplify to find the terms.


    STEP: Substitute y=3 into the expressions for the terms
    [−3 points ⇒ 0 / 3 points left]

    To find the values of the terms, we need to substitute y=3 into the expressions for each of the three terms:

    First term: T1=3y21=3(3)21=112Second term: T2=3y22=3(3)22=52Third term: T3=11y26=11(3)26=212

    Therefore, the first three terms of the sequence are: 112,52 and 212.


    Submit your answer as: andand

ID is: 2741 Seed is: 7386

Working with the common difference

  1. Compute the value of p for the following arithmetic sequence:

    3p+43;5p+133;5p+103;

    If the answer is a non-integer, write the answer as a simplified fraction.

    Answer: p=
    fraction
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    The key to starting this question is the fact that the terms form an arithmetic sequence: that means that the difference between the first and second terms is equal to the difference between the second and third terms.


    STEP: Set up an equation based on the common difference of the terms
    [−1 point ⇒ 3 / 4 points left]

    The given sequence is arithmetic, so the difference between successive terms will be the same (common). So we know that T2T1=T3T2 and we can solve for p:

    T2T1=T3T2(5p+133)(3p+43)=(5p+103)(5p+133)

    STEP: Simplify the equation
    [−2 points ⇒ 1 / 4 points left]

    For an equation with fractions, we can multiply both sides of the equation by the LCD of the fractions to remove the denominators. In this case, the LCD is 3.

    3(5p+133)3(3p+43)=3(5p+103)3(5p+133)15p+13(9p+4)=15p+10(15p+13)6p+9=3

    STEP: Solve for p
    [−1 point ⇒ 0 / 4 points left]

    Solve for p:

    12=6p 2=p

    Therefore, p=2.


    Submit your answer as:
  2. Determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions, not decimals.

    Answer: ; ; ;
    fraction
    fraction
    fraction
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The first three terms in the sequence are given in terms of p. We have calculated the value of p, so we can substitute and simplify to find the terms.


    STEP: Substitute p=2 into the expressions for the terms
    [−3 points ⇒ 0 / 3 points left]

    To find the values of the terms, we need to substitute p=2 into the expressions for each of the three terms:

    First term: T1=3p+43=3(2)+43=143Second term: T2=5p+133=5(2)+133=173Third term: T3=5p+103=5(2)+103=203

    Therefore, the first three terms of the sequence are: 143,173 and 203.


    Submit your answer as: andand

ID is: 3090 Seed is: 629

Arithmetic sequences with fractions

Consider the following arithmetic sequence:

10;545;585;625;
  1. Calculate the common difference (d).

    Answer: d=
    numeric
    2 attempts remaining
    STEP: Calculate the common difference (d).
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 4 terms of a sequence and specifically tells us that the numbers form an arithmetic pattern. We need to find the common difference of that pattern, d.

    To determine the common difference for an arithmetic sequence, we subtract any two successive terms in the sequence: d=TnTn1. In other words the common difference is the difference between any term in the sequence and the term before it:

    d=T2T1=(545)(10)=45

    Or we could also have chosen:

    d=T3T2=(585)(545)=45

    Notice that both calculations give the same value for d.

    Therefore the common difference is d=45.


    Submit your answer as:
  2. Find the next 2 terms in the sequence.

    Answer:

    T5=

    T6=

    numeric
    numeric
    2 attempts remaining
    STEP: Find the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    Now we must find the next 2 terms in the sequence. We can do this using the common difference: any term in the sequence is equal to the term before it plus the common difference: Tn=Tn1+d.

    To calculate the fifth term, add the common difference to the fourth term:

    T5=T4+(45)=(625)45=665

    Continue the sequence by adding d=45 again:

    T6=T5+(45)=(665)45=14

    Therefore, the next two terms in the sequence are: 665 and 14.

    The full sequence is 10;545;585;625;665;14;745;.


    Submit your answer as: and

ID is: 3090 Seed is: 4550

Arithmetic sequences with fractions

Consider the following arithmetic sequence:

7;112;4;52;1;12;
  1. Find the common difference, d.

    Answer: d=
    numeric
    2 attempts remaining
    STEP: Find the common difference, d.
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 6 terms of a sequence and specifically tells us that the numbers form an arithmetic pattern. We need to find the common difference of that pattern, d.

    To determine the common difference for an arithmetic sequence, we subtract any two successive terms in the sequence: d=TnTn1. In other words the common difference is the difference between any term in the sequence and the term before it:

    d=T2T1=(112)(7)=32

    Or we could also have chosen:

    d=T3T2=(4)(112)=32

    Notice that both calculations give the same value for d.

    Therefore the common difference is d=32.


    Submit your answer as:
  2. Find the next 2 terms in the sequence.

    Answer:

    T7=

    T8=

    numeric
    numeric
    2 attempts remaining
    STEP: Find the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    Now we must find the next 2 terms in the sequence. We can do this using the common difference: any term in the sequence is equal to the term before it plus the common difference: Tn=Tn1+d.

    To calculate the seventh term, add the common difference to the sixth term:

    T7=T6+(32)=(12)32=2

    Continue the sequence by adding d=32 again:

    T8=T7+(32)=(2)32=72

    Therefore, the next two terms in the sequence are: 2 and 72.

    The full sequence is 7;112;4;52;1;12;2;72;5;.


    Submit your answer as: and

ID is: 3090 Seed is: 3423

Arithmetic sequences with fractions

For the following arithmetic sequence:

6;275;245;215;
  1. Calculate the common difference (d).

    Answer: d=
    numeric
    2 attempts remaining
    STEP: Calculate the common difference (d).
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 4 terms of a sequence and specifically tells us that the numbers form an arithmetic pattern. We need to find the common difference of that pattern, d.

    To determine the common difference for an arithmetic sequence, we subtract any two successive terms in the sequence: d=TnTn1. In other words the common difference is the difference between any term in the sequence and the term before it:

    d=T2T1=(275)(6)=35

    Or we could also have chosen:

    d=T3T2=(245)(275)=35

    Notice that both calculations give the same value for d.

    Therefore the common difference is d=35.


    Submit your answer as:
  2. Find the next 2 terms in the sequence.

    Answer:

    T5=

    T6=

    numeric
    numeric
    2 attempts remaining
    STEP: Find the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    Now we must find the next 2 terms in the sequence. We can do this using the common difference: any term in the sequence is equal to the term before it plus the common difference: Tn=Tn1+d.

    To calculate the fifth term, add the common difference to the fourth term:

    T5=T4+(35)=(215)35=185

    Continue the sequence by adding d=35 again:

    T6=T5+(35)=(185)35=3

    Therefore, the next two terms in the sequence are: 185 and 3.

    The full sequence is 6;275;245;215;185;3;125;.


    Submit your answer as: and

ID is: 2689 Seed is: 1199

Common difference for an algebraic sequence

  1. Consider the following sequence:

    2p3;5p1;8p+1;11p+3;14p+5;

    Calculate the common difference (if there is one) for the terms of the sequence or write 'no common difference' in the answer box.

    Answer:
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The common difference for terms in an arithmetic pattern: d=Tn+1Tn.


    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the common difference d at least two different ways to make sure that the values are the same:

    d=T2T1=(5p1)(2p3)=3p+2d=T3T2=(8p+1)(5p1)=3p+2d=T4T3=(11p+3)(8p+1)=3p+2

    Each calculation gives the same answer, which means we have found the common difference: d=3p+2.


    Submit your answer as:
  2. If it is given that p=1, determine the values of T1 and T2.

    Answer: T1 = and T2 =
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    From the question we know that T1=2p3 and T2=5p1.
    STEP: Substitute the given value into the expression for each term.
    [−2 points ⇒ 0 / 2 points left]

    Substitute p=1 into the expressions for T1 and T2:

    T1=2p3=2(1)3=5T2=5p1=5(1)1=6

    Therefore, T1 = -5 and T2 = -6.


    Submit your answer as: and

ID is: 2689 Seed is: 4129

Common difference for an algebraic sequence

  1. Consider the following list:

    5y2;2y1;y;4y+1;7y+2;

    Calculate the common difference (if there is one) for the terms of the list or write 'no common difference' in the answer box.

    Answer:
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The common difference for terms in an arithmetic pattern: d=Tn+1Tn.


    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the common difference d at least two different ways to make sure that the values are the same:

    d=T2T1=(2y1)(5y2)=3y+1d=T3T2=(y)(2y1)=3y+1d=T4T3=(4y+1)(y)=3y+1

    Each calculation gives the same answer, which means we have found the common difference: d=3y+1.


    Submit your answer as:
  2. If it is given that y=1, determine the values of T1 and T2.

    Answer: T1 = and T2 =
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    From the question we know that T1=5y2 and T2=2y1.
    STEP: Substitute the given value into the expression for each term.
    [−2 points ⇒ 0 / 2 points left]

    Substitute y=1 into the expressions for T1 and T2:

    T1=5y2=5(1)2=3T2=2y1=2(1)1=1

    Therefore, T1 = 3 and T2 = 1.


    Submit your answer as: and

ID is: 2689 Seed is: 8203

Common difference for an algebraic sequence

  1. Consider the following list:

    3z+1;2z1;z3;5;z7;

    Find the common difference (if there is one) for the terms of the list or write 'no common difference' in the answer box.

    Answer:
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The common difference for terms in an arithmetic pattern: d=Tn+1Tn.


    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the common difference d at least two different ways to make sure that the values are the same:

    d=T2T1=(2z1)(3z+1)=z2d=T3T2=(z3)(2z1)=z2d=T4T3=(5)(z3)=z2

    Each calculation gives the same answer, which means we have found the common difference: d=z2.


    Submit your answer as:
  2. If it is given that z=3, determine the values of T1 and T2.

    Answer: T1 = and T2 =
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    From the question we know that T1=3z+1 and T2=2z1.
    STEP: Substitute the given value into the expression for each term.
    [−2 points ⇒ 0 / 2 points left]

    Substitute z=3 into the expressions for T1 and T2:

    T1=3z+1=3(3)+1=10T2=2z1=2(3)1=5

    Therefore, T1 = 10 and T2 = 5.


    Submit your answer as: and

ID is: 2796 Seed is: 4447

Finding the common difference

Calculate the common difference (if there is one) for the list:

2.59;3.91;5.23;6.55;

Give your answer as a decimal or write 'no common difference' in the answer box.

Answer: The common difference is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

We can check if there is a common difference between successive terms by calculating:

d=TnTn1

STEP: Calculate the difference between consecutive terms
[−1 point ⇒ 0 / 1 points left]

Use any two consecutive terms to calculate d. Remember: this may not be an arithmetic sequence, so we need to calculate d using at least two different pairs of terms to see whether or not the values are the same:

d=T2T1=3.91(2.59)=1.32d=T3T2=5.23(3.91)=1.32

In this case, we see that the two answers are the same, which means that the sequence of numbers is arithmetic. Therefore, the common difference is d=1.32.


Submit your answer as:

ID is: 2796 Seed is: 7081

Finding the common difference

Find the common difference (if there is one) for the list:

1.12;0.4;0.32;1.04;

Give your answer as a decimal or write 'no common difference' in the answer box.

Answer: The common difference is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

We can check if there is a common difference between successive terms by calculating:

d=TnTn1

STEP: Calculate the difference between consecutive terms
[−1 point ⇒ 0 / 1 points left]

Use any two consecutive terms to calculate d. Remember: this may not be an arithmetic sequence, so we need to calculate d using at least two different pairs of terms to see whether or not the values are the same:

d=T2T1=0.4(1.12)=0.72d=T3T2=0.32(0.4)=0.72

In this case, we see that the two answers are the same, which means that the sequence of numbers is arithmetic. Therefore, the common difference is d=0.72.


Submit your answer as:

ID is: 2796 Seed is: 9574

Finding the common difference

Find the common difference (if there is one) for the list:

0.01;1.01;2.01;3.01;

Give your answer as a decimal or write 'no common difference' in the answer box.

Answer: The common difference is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

We can check if there is a common difference between successive terms by calculating:

d=TnTn1

STEP: Calculate the difference between consecutive terms
[−1 point ⇒ 0 / 1 points left]

Use any two consecutive terms to calculate d. Remember: this may not be an arithmetic sequence, so we need to calculate d using at least two different pairs of terms to see whether or not the values are the same:

d=T2T1=1.01(0.01)=1.0d=T3T2=2.01(1.01)=1.0

In this case, we see that the two answers are the same, which means that the sequence of numbers is arithmetic. Therefore, the common difference is d=1.0.


Submit your answer as:

ID is: 2747 Seed is: 3540

Arithmetic sequences: finding the general formula

Determine the general formula and then calculate T10, T15 and T30 for the following sequences:

  1. 10;6;2;2;6;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    Find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Calculate the common difference:

    d=T2T1=6(10)=4Or d=T3T2=2(6)=4

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Determine the general formula:

    T1=a=10Tn=a+(n1)d=10+(n1)(4)Tn=4n14

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Find the three terms:

    T10=4(10)14=26T15=4(15)14=46T30=4(30)14=106

    Therefore, T10=26, T15=46 and T30=106.


    Submit your answer as: andandand
  2. 11;16;21;26;31;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    We follow the same steps as the first part of the question: find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Find the common difference:

    d=T2T1=16(11)=5

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now determine the general formula:

    T1=a=11Tn=a+(n1)d=11+(n1)(5)Tn=5n+6

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Finally, calculate the three required terms:

    T10=5(10)+6=56T15=5(15)+6=81T30=5(30)+6=156

    Therefore, T10=56, T15=81 and T30=156.


    Submit your answer as: andandand

ID is: 2747 Seed is: 7112

Arithmetic sequences: finding the general formula

Determine the general formula and then calculate T10, T15 and T30 for the following sequences:

  1. 8;6;4;2;0;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    Find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Calculate the common difference:

    d=T2T1=6(8)=2Or d=T3T2=4(6)=2

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Determine the general formula:

    T1=a=8Tn=a+(n1)d=8+(n1)(2)Tn=2n+10

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Find the three terms:

    T10=2(10)+10=10T15=2(15)+10=20T30=2(30)+10=50

    Therefore, T10=10, T15=20 and T30=50.


    Submit your answer as: andandand
  2. 24;33;42;51;60;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    We follow the same steps as the first part of the question: find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Find the common difference:

    d=T2T1=33(24)=9

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now determine the general formula:

    T1=a=24Tn=a+(n1)d=24+(n1)(9)Tn=9n+15

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Finally, calculate the three required terms:

    T10=9(10)+15=105T15=9(15)+15=150T30=9(30)+15=285

    Therefore, T10=105, T15=150 and T30=285.


    Submit your answer as: andandand

ID is: 2747 Seed is: 8114

Arithmetic sequences: finding the general formula

Determine the general formula and then calculate T10, T15 and T30 for the following sequences:

  1. 7;9;11;13;15;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    Find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Calculate the common difference:

    d=T2T1=9(7)=2Or d=T3T2=11(9)=2

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Determine the general formula:

    T1=a=7Tn=a+(n1)d=7+(n1)(2)Tn=2n5

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Find the three terms:

    T10=2(10)5=25T15=2(15)5=35T30=2(30)5=65

    Therefore, T10=25, T15=35 and T30=65.


    Submit your answer as: andandand
  2. 4;12;20;28;36;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    We follow the same steps as the first part of the question: find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Find the common difference:

    d=T2T1=12(4)=8

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now determine the general formula:

    T1=a=4Tn=a+(n1)d=4+(n1)(8)Tn=8n4

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Finally, calculate the three required terms:

    T10=8(10)4=76T15=8(15)4=116T30=8(30)4=236

    Therefore, T10=76, T15=116 and T30=236.


    Submit your answer as: andandand

2. Arithmetic sequences

3. Arithmetic series


ID is: 3926 Seed is: 3833

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3
Maths formulas

Consider the series:

Sn=3+3+9+15+

to n terms.

  1. We can find an expression for the general term of the series in the form Tk=bk+c. What are the values of b and c?

    Answer:
    1. b=
    2. c=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    First find the values of a and d for this series.


    STEP: Determine the values of a and d
    [−1 point ⇒ 1 / 2 points left]

    The formula for the general term of an arithmetic series is

    Tn=a+(n1)d

    a is the first term of the series. So here a=3.

    d is the constant difference between the terms of the series. We can determine this by finding the difference between any two consecutive terms of the series. Taking the difference of the first two terms:

    d=T2T1=3(3)=6

    STEP: Substitute into the formula and simplify
    [−1 point ⇒ 0 / 2 points left]

    To find Tk, we can now substitute the values of a and d into the formula for the general term and simplify.

    Tk=(3)+(k1)(6)=3+6k6=6k9

    We were looking for b and c in the expression Tk=bk+c. So the correct answers are

    1. b=6
    2. c=9

    Submit your answer as: and
  2. Write Sn in sigma notation, and then answer the two questions below about the expression you find. It should look something like

    Sn=xy(Tk)

    What symbols or expressions should be in the place of x and y above?

    Answer:
    1. x:
    2. y:
    equation
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise sigma notation in the Everything Maths textbook.


    STEP: Write Sn in sigma notation and read off the information you need
    [−2 points ⇒ 0 / 2 points left]

    Writing Sn in sigma notation, we get:

    Sn=k=1n(6k9)

    So the correct answers are

    1. x:k=1
    2. y:n

    Submit your answer as: and
  3. Derive an algebraic expression for Sn.

    INSTRUCTION: Simplify your answer completely.
    Answer: Sn=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute the values of a and d into the formula for the sum of the series.


    STEP: Substitute a and d into the formula for Sn
    [−3 points ⇒ 0 / 3 points left]

    The formula for the sum of n terms in an arithmetic series is:

    Sn=n2[2a+(n1)d]

    We need to substitute the values of a and d into this formula, and simplify.

    Sn=n2[2a+(n1)d]=n2[2(3)+(n1)(6)]=n2[6+6n6]=n2[6n12]=3n26n

    Submit your answer as:

ID is: 3926 Seed is: 3614

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3
Maths formulas

Consider the series:

Sn=42814

to n terms.

  1. We can find an expression for the general term of the series in the form Tk=bk+c. What are the values of b and c?

    Answer:
    1. b=
    2. c=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    First find the values of a and d for this series.


    STEP: Determine the values of a and d
    [−1 point ⇒ 1 / 2 points left]

    The formula for the general term of an arithmetic series is

    Tn=a+(n1)d

    a is the first term of the series. So here a=4.

    d is the constant difference between the terms of the series. We can determine this by finding the difference between any two consecutive terms of the series. Taking the difference of the first two terms:

    d=T2T1=2(4)=6

    STEP: Substitute into the formula and simplify
    [−1 point ⇒ 0 / 2 points left]

    To find Tk, we can now substitute the values of a and d into the formula for the general term and simplify.

    Tk=(4)+(k1)(6)=46k+6=6k+10

    We were looking for b and c in the expression Tk=bk+c. So the correct answers are

    1. b=6
    2. c=10

    Submit your answer as: and
  2. Write Sn in sigma notation, and then answer the two questions below about the expression you find. It should look something like

    Sn=xy(Tk)

    What symbols or expressions should be in the place of x and y above?

    Answer:
    1. x:
    2. y:
    equation
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise sigma notation in the Everything Maths textbook.


    STEP: Write Sn in sigma notation and read off the information you need
    [−2 points ⇒ 0 / 2 points left]

    Writing Sn in sigma notation, we get:

    Sn=k=1n(6k+10)

    So the correct answers are

    1. x:k=1
    2. y:n

    Submit your answer as: and
  3. Derive an algebraic expression for Sn.

    INSTRUCTION: Simplify your answer completely.
    Answer: Sn=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute the values of a and d into the formula for the sum of the series.


    STEP: Substitute a and d into the formula for Sn
    [−3 points ⇒ 0 / 3 points left]

    The formula for the sum of n terms in an arithmetic series is:

    Sn=n2[2a+(n1)d]

    We need to substitute the values of a and d into this formula, and simplify.

    Sn=n2[2a+(n1)d]=n2[2(4)+(n1)(6)]=n2[86n+6]=n2[6n+14]=3n2+7n

    Submit your answer as:

ID is: 3926 Seed is: 357

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3
Maths formulas

Consider the series:

Sn=261422

to n terms.

  1. We can find an expression for the general term of the series in the form Tk=bk+c. What are the values of b and c?

    Answer:
    1. b=
    2. c=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    First find the values of a and d for this series.


    STEP: Determine the values of a and d
    [−1 point ⇒ 1 / 2 points left]

    The formula for the general term of an arithmetic series is

    Tn=a+(n1)d

    a is the first term of the series. So here a=2.

    d is the constant difference between the terms of the series. We can determine this by finding the difference between any two consecutive terms of the series. Taking the difference of the first two terms:

    d=T2T1=6(2)=8

    STEP: Substitute into the formula and simplify
    [−1 point ⇒ 0 / 2 points left]

    To find Tk, we can now substitute the values of a and d into the formula for the general term and simplify.

    Tk=(2)+(k1)(8)=28k+8=8k+10

    We were looking for b and c in the expression Tk=bk+c. So the correct answers are

    1. b=8
    2. c=10

    Submit your answer as: and
  2. Write Sn in sigma notation, and then answer the two questions below about the expression you find. It should look something like

    Sn=xy(Tk)

    What symbols or expressions should be in the place of x and y above?

    Answer:
    1. x:
    2. y:
    equation
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise sigma notation in the Everything Maths textbook.


    STEP: Write Sn in sigma notation and read off the information you need
    [−2 points ⇒ 0 / 2 points left]

    Writing Sn in sigma notation, we get:

    Sn=k=1n(8k+10)

    So the correct answers are

    1. x:k=1
    2. y:n

    Submit your answer as: and
  3. Derive an algebraic expression for Sn.

    INSTRUCTION: Simplify your answer completely.
    Answer: Sn=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute the values of a and d into the formula for the sum of the series.


    STEP: Substitute a and d into the formula for Sn
    [−3 points ⇒ 0 / 3 points left]

    The formula for the sum of n terms in an arithmetic series is:

    Sn=n2[2a+(n1)d]

    We need to substitute the values of a and d into this formula, and simplify.

    Sn=n2[2a+(n1)d]=n2[2(2)+(n1)(8)]=n2[48n+8]=n2[8n+12]=4n2+6n

    Submit your answer as:

ID is: 3927 Seed is: 7145

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3.4
Maths formulas

A series is defined as

Sn=2+2+6+10+

to n terms. Sn is also described by the expression

Sn=2n24n

Another sequence is defined as:

Q1=7Q2=72Q3=72+2Q4=72+2+6Q5=72+2+6+10
  1. Complete the expression below for Q6.

    Answer: Q6= -7 - 2 + 2 + 6 + +
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]


    STEP: Fill in the missing values in Q6
    [−2 points ⇒ 0 / 2 points left]

    Taking a closer look at the terms of Q, we can see that each new term of Q is the same as the term before it, with one extra number at the end.

    Q1=7 is a random number. But each of the other numbers which are being added to Q is a term of Sn. For instance:

    Q2=72=7+S1

    and

    Q3=72+2=7+S2

    and so on. So the general term for Qn is

    Qn=7+Sn1

    Sn has a constant difference of d=4. So we can use this to find the missing values of Q6.

    Q6=72+2+6+10+14

    Submit your answer as: and
  2. Calculate the value of Q111.

    Answer: Q111=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the fact that you know Qn in terms of S, and that you are given an algebraic expression for Sn.


    STEP: Substitute the expression for Sn into Q111 and simplify
    [−3 points ⇒ 0 / 3 points left]

    From Question 1, we know that

    Qn=7+Sn1

    And we are given in the question that

    Sn=2n24n

    We can use these together to find the value of Q111.

    Q111=7+S110=7+2(110)24(110)=7+24200440=23753

    Submit your answer as:

ID is: 3927 Seed is: 4677

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3.4
Maths formulas

A series is defined as

Sn=2+2+6+10+

to n terms. Sn is also described by the expression

Sn=2n24n

Another sequence is defined as:

Q1=9Q2=92Q3=92+2Q4=92+2+6Q5=92+2+6+10
  1. Complete the expression below for Q7.

    Answer: Q7= 9 2 + 2 + 6 + 10 + +
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]


    STEP: Fill in the missing values in Q7
    [−2 points ⇒ 0 / 2 points left]

    Taking a closer look at the terms of Q, we can see that each new term of Q is the same as the term before it, with one extra number at the end.

    Q1=9 is a random number. But each of the other numbers which are being added to Q is a term of Sn. For instance:

    Q2=92=9+S1

    and

    Q3=92+2=9+S2

    and so on. So the general term for Qn is

    Qn=9+Sn1

    Sn has a constant difference of d=4. So we can use this to find the missing values of Q7.

    Q7=92+2+6+10+14+18

    Submit your answer as: and
  2. Calculate the value of Q121.

    Answer: Q121=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the fact that you know Qn in terms of S, and that you are given an algebraic expression for Sn.


    STEP: Substitute the expression for Sn into Q121 and simplify
    [−3 points ⇒ 0 / 3 points left]

    From Question 1, we know that

    Qn=9+Sn1

    And we are given in the question that

    Sn=2n24n

    We can use these together to find the value of Q121.

    Q121=9+S120=9+2(120)24(120)=9+28800480=28311

    Submit your answer as:

ID is: 3927 Seed is: 7758

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3.4
Maths formulas

A series is defined as

Sn=441220

to n terms. Sn is also described by the expression

Sn=4n2+8n

Another sequence is defined as:

Q1=7Q2=7+4Q3=7+44Q4=7+4412Q5=7+441220
  1. Complete the expression below for Q7.

    Answer: Q7= 7 + 4 4 12 20
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]


    STEP: Fill in the missing values in Q7
    [−2 points ⇒ 0 / 2 points left]

    Taking a closer look at the terms of Q, we can see that each new term of Q is the same as the term before it, with one extra number at the end.

    Q1=7 is a random number. But each of the other numbers which are being added to Q is a term of Sn. For instance:

    Q2=7+4=7+S1

    and

    Q3=7+44=7+S2

    and so on. So the general term for Qn is

    Qn=7+Sn1

    Sn has a constant difference of d=8. So we can use this to find the missing values of Q7.

    Q7=7+4412202836

    Submit your answer as: and
  2. Calculate the value of Q115.

    Answer: Q115=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the fact that you know Qn in terms of S, and that you are given an algebraic expression for Sn.


    STEP: Substitute the expression for Sn into Q115 and simplify
    [−3 points ⇒ 0 / 3 points left]

    From Question 1, we know that

    Qn=7+Sn1

    And we are given in the question that

    Sn=4n2+8n

    We can use these together to find the value of Q115.

    Q115=7+S114=74(114)2+8(114)=751984+912=51065

    Submit your answer as:

ID is: 2939 Seed is: 4125

Finding the sum for a finite arithmetic series

  1. The first term of an arithmetic series is 10 while the last term is 70. In total, there are 17 terms in the series. Find the sum of these 17 terms, S17.

    Answer: S17=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(a+l).


    STEP: Identify the values of n,a and l
    [−1 point ⇒ 2 / 3 points left]

    For this question we have the first and last terms of an arithmetic series, and we need to calculate the sum of the first 17 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(a+l):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • l is the last term of the series

    We need the values of n,a and l - then we can substitute them into the formula and evaluate the answer.

    We can take the value of n straight from the question: it is in the words "the sum of these 17 terms" as well as in the symbol "S17." Since we want to add up 17 terms of the series, n=17.

    The first term is given in the question: a=10. We can also get the last term in the series directly from the question: l=70.

    We now have the values n=17,a=10 and l=70.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(a+l)S17=172((10)+(70))=172(60)=510

    The sum of the first 17 terms for the series is S17=510.


    Submit your answer as:
  2. Based on the first question, find the sum of the first 18 terms of the series, S18 , if the term following the last term is T18=75.

    Answer: S18= 
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the sum S18. Since the question is only worth one mark, we can expect that there is a quick way (faster than the calculation in the first question).

    Compare this question to the first one: in the first question we found the sum S17. Think about what this means: it is the sum of the first 17 terms, as shown below.

    S17=T1+T2+T3++T16+T17

    Now we want to find the sum of the first 18 terms, which is the same as the sum above, except with one more term added on! In other words, S18=S17+T18. We can substitute the values we know into this equation to get the answer.

    S18=S17+T18=(510)+(75)=585

    The sum of the first 18 terms for the series is S18=585.


    Submit your answer as:

ID is: 2939 Seed is: 7481

Finding the sum for a finite arithmetic series

  1. The first 27 terms of an arithmetic series are 4+(3)+(2)++22. Compute the sum of these 27 terms for the series, S27.

    Answer: S27=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(a+l).


    STEP: Identify the values of n,a and l
    [−1 point ⇒ 2 / 3 points left]

    For this question we have the first three terms and the last term of an arithmetic series, and we need to calculate the sum of the first 27 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(a+l):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • l is the last term of the series

    We need the values of n,a and l - then we can substitute them into the formula and evaluate the answer.

    We can take the value of n straight from the question: it is in the words "the sum of these 27 terms" as well as in the symbol "S27." Since we want to add up 27 terms of the series, n=27.

    The first term is given in the question: a=4. The last term is also given in the question l=22. (The question also shows the second and third terms, but we do not need them to answer the question.)

    We now have the values n=27,a=4 and l=22.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(a+l)S27=272((4)+(22))=272(18)=243

    The sum of the first 27 terms for the series is S27=243.


    Submit your answer as:
  2. Based on the first question, find the sum of the first 26 terms of the series, S26 . Remember that the last term, given above, has a value of T27=22.

    Answer: S26= 
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the sum S26. Since the question is only worth one mark, we can expect that there is a quick way (faster than the calculation in the first question).

    Compare this question to the first one: in the first question we found the sum S27. Think about what this means: it is the sum of the first 27 terms, as shown below.

    S27=T1+T2+T3++T26+T27

    Now we want to find the sum of the first 26 terms, which is the same as above except with the last term taken away! In other words, S26=S27T27. We can substitute the values we know into this equation to get the answer.

    S26=S27T27=(243)(22)=221

    The sum of the first 26 terms for the series is S26=221.


    Submit your answer as:

ID is: 2939 Seed is: 9610

Finding the sum for a finite arithmetic series

  1. The first term of an arithmetic series is 9 while the last term is 97. In total, there are 23 terms in the series. Determine the sum of these 23 terms, S23.

    Answer: S23=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(a+l).


    STEP: Identify the values of n,a and l
    [−1 point ⇒ 2 / 3 points left]

    For this question we have the first and last terms of an arithmetic series, and we need to calculate the sum of the first 23 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(a+l):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • l is the last term of the series

    We need the values of n,a and l - then we can substitute them into the formula and evaluate the answer.

    We can take the value of n straight from the question: it is in the words "the sum of these 23 terms" as well as in the symbol "S23." Since we want to add up 23 terms of the series, n=23.

    The first term is given in the question: a=9. We can also get the last term in the series directly from the question: l=97.

    We now have the values n=23,a=9 and l=97.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(a+l)S23=232((9)+(97))=232(106)=1,219

    The sum of the first 23 terms for the series is S23=1219.


    Submit your answer as:
  2. Based on the first question, find the sum of the first 22 terms of the series, S22 . Remember that the last term, given above, has a value of T23=97.

    Answer: S22= 
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the sum S22. Since the question is only worth one mark, we can expect that there is a quick way (faster than the calculation in the first question).

    Compare this question to the first one: in the first question we found the sum S23. Think about what this means: it is the sum of the first 23 terms, as shown below.

    S23=T1+T2+T3++T22+T23

    Now we want to find the sum of the first 22 terms, which is the same as above except with the last term taken away! In other words, S22=S23T23. We can substitute the values we know into this equation to get the answer.

    S22=S23T23=(1219)(97)=1122

    The sum of the first 22 terms for the series is S22=1122.


    Submit your answer as:

ID is: 2937 Seed is: 8781

Working with arithmetic series

  1. The difference between the third and sixth terms of an arithmetic series is -24 and the sum of the first 38 terms is -5396. Find the first three terms in the series.

    Answer:

    T1=

    T2=

    T3=

    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    Use the equation Sn=n2(2a+(n1)d) to write an equation using the information about the sum of the first 38 terms. Then write out another equation based on the difference between the sixth and third terms.


    STEP: Use the formula Sn=n2(2a+(n1)d) to write an equation for the sum of the first 38 terms
    [−1 point ⇒ 5 / 6 points left]

    There is a lot of information in the question. We will start with the statement that the sum of the first 38 terms is -5396. This information fits with the formula Sn=n2(2a+(n1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.

    Substitute in what we know to see what the equation can give us:

    Sn=n2(2a+(n1)d)5396=382(2a+(381)d)10792=38(2a+37d)284=2a+37d

    For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.


    STEP: Use the information about the other terms to write another equation
    [−2 points ⇒ 3 / 6 points left]

    In the work above, we got an equation with two variables, so the plan for the question is now clear: find another equation with the variables a and d so that we can solve the equations simultaneously. Let us try to do that using the information in the question about the other terms.

    In this case, the question states that the "difference between the third and sixth terms ... is -24". We can write this as an equation: T6T3=24. Using the relationship Tn=a+(n1)d, we can write this equation in terms of a and d:

    T6T3=24(a+5d)(a+2d)=243d=24d=8

    This is fantastic: the a's cancelled so we could actually find the value of the common difference, d.


    STEP: Use the first equation and the value of d to find a
    [−1 point ⇒ 2 / 6 points left]

    We now have the two equations:

    2a+37d=284(1)d=8(2)

    Substitute d=8 into equation (1) to find the value of a.

    2a+37d=2842a+37(8)=2842a=12a=6

    We now have the value of the first term, T1=a=6, and the common difference, d=8.


    STEP: Calculate the values of the second and third terms
    [−2 points ⇒ 0 / 6 points left]

    Now use the values of a=6 and d=8 to calculate the values of T2 and T3.

    T2=a+d=6+(8)=2T3=a+2d=6+2(8)=10

    The first three terms of the sequence are: T1=6,T2=2,T3=10.


    Submit your answer as: andand
  2. Hence, if you are now told that S39 is equal to -5694, calculate the value of T39.

    Answer:

    T39=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can find the answer by comparing the sums S39 and S38.


    STEP: Compare the sums S39 and S38 to find the value of T39
    [−1 point ⇒ 0 / 1 points left]

    Sn is the sum of n terms of a series: Sn=T1+T2+T3++Tn. In this question we have S39 and S38:

    S39=T1+T2+T3++T37+T38+T39S38=T1+T2+T3++T37+T38

    As you can see, the only difference between S39 and S38 is one term, T39. If we subtract T39 from S39 we will get S38. In other words, T39 is the difference between S39 and S38:

    T39=S39S38

    Now we can substitute in the values of the two sums (S39=5694 and S38=5396) and find the term we want:

    T39=S39S38=(5694)(5396)=298

    The correct answer is T39=298.


    Submit your answer as:

ID is: 2937 Seed is: 5558

Working with arithmetic series

  1. The sum of the first 28 terms in an arithmetic series is -1638. The series includes the terms T4=6 and T9=31. Find the first three terms in the series.

    Answer:

    T1=

    T2=

    T3=

    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    Use the equation Sn=n2(2a+(n1)d) to write an equation using the information about the sum of the first 28 terms. Then write out another equation based on the either T4 or T9.


    STEP: Use the formula Sn=n2(2a+(n1)d) to write an equation for the sum of the first 28 terms
    [−1 point ⇒ 5 / 6 points left]

    There is a lot of information in the question. We will start with the statement that the sum of the first 28 terms is -1638. This information fits with the formula Sn=n2(2a+(n1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.

    Substitute in what we know to see what the equation can give us:

    Sn=n2(2a+(n1)d)1638=282(2a+(281)d)3276=28(2a+27d)117=2a+27d

    For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.


    STEP: Use the information about the other terms to write another equation
    [−1 point ⇒ 4 / 6 points left]

    In the work above, we got an equation with two variables, so the plan for the question is now clear: find another equation with the variables a and d so that we can solve the equations simultaneously. Let us try to do that using the information in the question about the other terms.

    In this case, we have the values for two different terms, T4 and T9. We can use either one of them to write an equation using the formula Tn=a+(n1)d. We will use T4=6.

    Tn=a+(n1)d6=a+(41)d6=a+3d


    STEP: Solve the two equations simultaneously for a and d
    [−2 points ⇒ 2 / 6 points left]

    We now have the two equations:

    2a+27d=117(1)a+3d=6(2)

    Now we can solve the equations simultaneously. Multiply equation (2) by 2 to get 2a+6d=12: that way both equations include 2a and we can use elimination.

    2a+27d=117(1)2a+6d=12(2)Eqn (1) - (2):21d=105d=5 And2a+6(5)=122a=18a=9

    We now have the value of the first term, T1=a=9, and the common difference, d=5.


    STEP: Calculate the values of the second and third terms
    [−2 points ⇒ 0 / 6 points left]

    Now use the values of a=9 and d=5 to calculate the values of T2 and T3.

    T2=a+d=9+(5)=4T3=a+2d=9+2(5)=1

    The first three terms of the sequence are: T1=9,T2=4,T3=1.


    Submit your answer as: andand
  2. Hence, if you are now told that S27 is equal to -1512, determine the value of T28.

    Answer:

    T28=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can find the answer by comparing the sums S28 and S27.


    STEP: Compare the sums S28 and S27 to find the value of T28
    [−1 point ⇒ 0 / 1 points left]

    Sn is the sum of n terms of a series: Sn=T1+T2+T3++Tn. In this question we have S28 and S27:

    S28=T1+T2+T3++T26+T27+T28S27=T1+T2+T3++T26+T27

    As you can see, the only difference between S28 and S27 is one term, T28. If we subtract T28 from S28 we will get S27. In other words, T28 is the difference between S28 and S27:

    T28=S28S27

    Now we can substitute in the values of the two sums (S28=1638 and S27=1512) and find the term we want:

    T28=S28S27=(1638)(1512)=126

    The correct answer is T28=126.


    Submit your answer as:

ID is: 2937 Seed is: 7545

Working with arithmetic series

  1. The difference between the second and fourth terms of an arithmetic series is -6 and the sum of the first 28 terms is -1022. Determine the first three terms in the series.

    Answer:

    T1=

    T2=

    T3=

    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    Use the equation Sn=n2(2a+(n1)d) to write an equation using the information about the sum of the first 28 terms. Then write out another equation based on the difference between the fourth and second terms.


    STEP: Use the formula Sn=n2(2a+(n1)d) to write an equation for the sum of the first 28 terms
    [−1 point ⇒ 5 / 6 points left]

    There is a lot of information in the question. We will start with the statement that the sum of the first 28 terms is -1022. This information fits with the formula Sn=n2(2a+(n1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.

    Substitute in what we know to see what the equation can give us:

    Sn=n2(2a+(n1)d)1022=282(2a+(281)d)2044=28(2a+27d)73=2a+27d

    For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.


    STEP: Use the information about the other terms to write another equation
    [−2 points ⇒ 3 / 6 points left]

    In the work above, we got an equation with two variables, so the plan for the question is now clear: find another equation with the variables a and d so that we can solve the equations simultaneously. Let us try to do that using the information in the question about the other terms.

    In this case, the question states that the "difference between the second and fourth terms ... is -6". We can write this as an equation: T4T2=6. Using the relationship Tn=a+(n1)d, we can write this equation in terms of a and d:

    T4T2=6(a+3d)(a+1d)=62d=6d=3

    This is fantastic: the a's cancelled so we could actually find the value of the common difference, d.


    STEP: Use the first equation and the value of d to find a
    [−1 point ⇒ 2 / 6 points left]

    We now have the two equations:

    2a+27d=73(1)d=3(2)

    Substitute d=3 into equation (1) to find the value of a.

    2a+27d=732a+27(3)=732a=8a=4

    We now have the value of the first term, T1=a=4, and the common difference, d=3.


    STEP: Calculate the values of the second and third terms
    [−2 points ⇒ 0 / 6 points left]

    Now use the values of a=4 and d=3 to calculate the values of T2 and T3.

    T2=a+d=4+(3)=1T3=a+2d=4+2(3)=2

    The first three terms of the sequence are: T1=4,T2=1,T3=2.


    Submit your answer as: andand
  2. Hence, if you are now told that S29 is equal to -1102, find the value of T29.

    Answer:

    T29=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can find the answer by comparing the sums S29 and S28.


    STEP: Compare the sums S29 and S28 to find the value of T29
    [−1 point ⇒ 0 / 1 points left]

    Sn is the sum of n terms of a series: Sn=T1+T2+T3++Tn. In this question we have S29 and S28:

    S29=T1+T2+T3++T27+T28+T29S28=T1+T2+T3++T27+T28

    As you can see, the only difference between S29 and S28 is one term, T29. If we subtract T29 from S29 we will get S28. In other words, T29 is the difference between S29 and S28:

    T29=S29S28

    Now we can substitute in the values of the two sums (S29=1102 and S28=1022) and find the term we want:

    T29=S29S28=(1102)(1022)=80

    The correct answer is T29=80.


    Submit your answer as:

ID is: 2948 Seed is: 5847

Finding the sum for a finite arithmetic series

  1. The first three terms of an arithmetic series are 4;9;14. Determine the sum of the first 12 terms of the series, S12.

    Answer: S12=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(2a+(n1)d).


    STEP: Identify the values of n,a and d
    [−2 points ⇒ 2 / 4 points left]

    The question gives us the first three terms of an arithmetic series, and we need to calculate the sum of the first 12 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(2a+(n1)d):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • d is the common difference in the series

    We need the values of n,a and d - then we can substitute them into the formula and evaluate the answer.

    The value of n is sitting in the question: we can find it in the words "the sum of the first 12 terms" as well as in the symbol "S12." For this question, we want to add up 12 terms of the series, so n=12.

    The first term is given in the question: a=4. The common difference is not given so clearly in the question - we must calculate it by using the terms given: d=T2T1=94=5.

    We now have the values n=12,a=4 and d=5.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 4 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(2a+(n1)d)S12=122(2(4)+(121)(5))=6(8+(11)(5))=6(8+55)=6(63)=378

    The sum of the first 12 terms for the series is S12=378.


    Submit your answer as:
  2. Use the answer to the first question to find the sum of the first 11 terms of the series, S11 , if the last term has a value of T12=59.

    Answer: S11= 
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We now have the new information T12=59 and need to determine the sum S11. One option is to use the formula again, as in the first question. However, there is a much quicker way (and the question is only worth one mark).

    Compare this question to the first one: in the first question we found the sum of the first 12 terms, as shown below.

    S12=T1+T2+T3++T11+T12

    Now we want to find the sum of the first 11 terms, which is the same as above except with the last term taken away! In other words, S11=S12T12. We can substitute the values we know into this equation to get the answer.

    S11=S12T12=(378)(59)=319

    The sum of the first 11 terms for the series is S11=319.


    Submit your answer as:

ID is: 2948 Seed is: 2790

Finding the sum for a finite arithmetic series

  1. The first term of an arithmetic series is 8 while the common difference is 3. Find the sum of the first 18 terms, S18.

    Answer: S18=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(2a+(n1)d).


    STEP: Identify the values of n,a and d
    [−1 point ⇒ 2 / 3 points left]

    The question gives us the first and common difference of an arithmetic series, and we need to calculate the sum of the first 18 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(2a+(n1)d):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • d is the common difference in the series

    We need the values of n,a and d - then we can substitute them into the formula and evaluate the answer.

    The value of n is sitting in the question: we can find it in the words "the sum of the first 18 terms" as well as in the symbol "S18." For this question, we want to add up 18 terms of the series, so n=18.

    The first term is given in the question: a=8. The common difference is also clearly stated in the question: d=3.

    We now have the values n=18,a=8 and d=3.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(2a+(n1)d)S18=182(2(8)+(181)(3))=9((16)+(17)(3))=9((16)+51)=9(35)=315

    The sum of the first 18 terms for the series is S18=315.


    Submit your answer as:
  2. Use the answer to the first question to find the sum of the first 17 terms of the series, S17 , if the last term has a value of T18=43.

    Answer: S17= 
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We now have the new information T18=43 and need to determine the sum S17. One option is to use the formula again, as in the first question. However, there is a much quicker way (and the question is only worth one mark).

    Compare this question to the first one: in the first question we found the sum of the first 18 terms, as shown below.

    S18=T1+T2+T3++T17+T18

    Now we want to find the sum of the first 17 terms, which is the same as above except with the last term taken away! In other words, S17=S18T18. We can substitute the values we know into this equation to get the answer.

    S17=S18T18=(315)(43)=272

    The sum of the first 17 terms for the series is S17=272.


    Submit your answer as:

ID is: 2948 Seed is: 4378

Finding the sum for a finite arithmetic series

  1. The first three terms of an arithmetic series are 7;10;13. Determine the sum of the first 16 terms of the series, S16.

    Answer: S16=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(2a+(n1)d).


    STEP: Identify the values of n,a and d
    [−2 points ⇒ 2 / 4 points left]

    The question gives us the first three terms of an arithmetic series, and we need to calculate the sum of the first 16 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(2a+(n1)d):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • d is the common difference in the series

    We need the values of n,a and d - then we can substitute them into the formula and evaluate the answer.

    The value of n is sitting in the question: we can find it in the words "the sum of the first 16 terms" as well as in the symbol "S16." For this question, we want to add up 16 terms of the series, so n=16.

    The first term is given in the question: a=7. The common difference is not given so clearly in the question - we must calculate it by using the terms given: d=T2T1=(10)(7)=3.

    We now have the values n=16,a=7 and d=3.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 4 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(2a+(n1)d)S16=162(2(7)+(161)(3))=8((14)+(15)(3))=8((14)+(45))=8(59)=472

    The sum of the first 16 terms for the series is S16=472.


    Submit your answer as:
  2. Use the answer to the first question to find the sum of the first 15 terms of the series, S15 , if the last term has a value of T16=52.

    Answer: S15= 
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We now have the new information T16=52 and need to determine the sum S15. One option is to use the formula again, as in the first question. However, there is a much quicker way (and the question is only worth one mark).

    Compare this question to the first one: in the first question we found the sum of the first 16 terms, as shown below.

    S16=T1+T2+T3++T15+T16

    Now we want to find the sum of the first 15 terms, which is the same as above except with the last term taken away! In other words, S15=S16T16. We can substitute the values we know into this equation to get the answer.

    S15=S16T16=(472)(52)=420

    The sum of the first 15 terms for the series is S15=420.


    Submit your answer as:

ID is: 3858 Seed is: 1841

Finite arithmetic sequence

Adapted from DBE Nov 2016 Grade 12, P1, Q2
Maths formulas

Given the finite arithmetic sequence:

4,1,2,,53,56
  1. Write down the sixth term (T6) of the sequence.

    Answer: T6=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Determine the value of the sixth term
    [−1 point ⇒ 0 / 1 points left]

    We can see from the first three terms that the common difference value for this sequence is 3. So the sixth term can be calculated as:

    Tn=a+(n1)dT6=(4)+((6)1)(3)=11

    Submit your answer as:
  2. Calculate the number of terms in the sequence.

    Answer: n=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Calculate the number of terms in the sequence
    [−3 points ⇒ 0 / 3 points left]

    We can use the same formula that we used in Question 1 to find the number of terms in the sequence, because we know the value of the final term in the sequence:

    Tn=a+(n1)d56=(4)+(n1)(3)56=4+3n363=3nn=21

    Submit your answer as:
  3. Calculate the sum of all the positive numbers in the sequence.

    Answer: The sum of the positive numbers is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Calculate the sum of the positive terms
    [−3 points ⇒ 0 / 3 points left]

    The first two terms are negative, and then all of the rest of the terms are positive. So we need to exclude the first two terms when we determine the sum of the positive terms.

    So the positive terms of the sequence look like:

    2,5,,56

    There will be 19 terms in this sequence, since we are excluding two negative terms. And the new value of a will be 2. We can then calculate the sum of these terms using the formula for the sum of an arithmetic series:

    Sn=n2[2a+(n1)d]S19=192[2(2)+((19)1)(3)]=38+541.528.5=551

    Submit your answer as:
  4. Consider the sequence

    4,1,2,,53,56,,2,873

    Determine the number of terms in this sequence that will be exactly divisible by 4.

    Answer: The number of terms divisible by 4 is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Determine the last term in the sequence divisible by 4
    [−0 points ⇒ 4 / 4 points left]

    The first term in the sequence is 4, which is divisible by 4.

    But the last term in the sequence is 2,873, which is not divisible by 4. So we need to work out the value of the last term which is divisible by 4.

    Since the common difference of the sequence is 3, we can count back from 2,873 in steps of 3 until we reach a number which is divisible by 4.

    TIP: Use your calculator if necessary to find the last value which is divisible by 4.
    2,8732,8702,8672,864

    So the last term which is divisible by 4 is 2,864.


    STEP: Continue the sequence to identify terms divisible by 4
    [−0 points ⇒ 4 / 4 points left]

    The original sequence is:

    4,1,2,5,8,11,14,17,20,,2,864

    The numbers divisible by 4 form a new sequence: 4,8,20,,2,864

    The new pattern is an arithmetic sequence with a=4 and d=12.


    STEP: Use the formula for arithmetic sequences to calculate the number of terms divisible by 4
    [−4 points ⇒ 0 / 4 points left]
    Tn=a+(n1)d2,864=4+(n1)(12)2,864=4+12n122,880=12n240=n

    Submit your answer as:

ID is: 3858 Seed is: 8711

Finite arithmetic sequence

Adapted from DBE Nov 2016 Grade 12, P1, Q2
Maths formulas

Given the finite arithmetic sequence:

5,1,3,,91,95
  1. Write down the fifth term (T5) of the sequence.

    Answer: T5=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Determine the value of the fifth term
    [−1 point ⇒ 0 / 1 points left]

    We can see from the first three terms that the common difference value for this sequence is 4. So the fifth term can be calculated as:

    Tn=a+(n1)dT5=(5)+((5)1)(4)=11

    Submit your answer as:
  2. Calculate the number of terms in the sequence.

    Answer: n=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Calculate the number of terms in the sequence
    [−3 points ⇒ 0 / 3 points left]

    We can use the same formula that we used in Question 1 to find the number of terms in the sequence, because we know the value of the final term in the sequence:

    Tn=a+(n1)d95=(5)+(n1)(4)95=54n+4104=4nn=26

    Submit your answer as:
  3. Calculate the sum of all the negative numbers in the sequence.

    Answer: The sum of the negative numbers is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Calculate the sum of the negative terms
    [−3 points ⇒ 0 / 3 points left]

    The first two terms are positive, and then all of the rest of the terms are negative. So we need to exclude the first two terms when we determine the sum of the negative terms.

    So the negative terms of the sequence look like:

    3,7,,95

    There will be 24 terms in this sequence, since we are excluding two positive terms. And the new value of a will be 3. We can then calculate the sum of these terms using the formula for the sum of an arithmetic series:

    Sn=n2[2a+(n1)d]S24=242[2(3)+((24)1)(4)]=721,152.0+48.0=1176

    Submit your answer as:
  4. Consider the sequence

    5,1,3,,91,95,,4,991

    Determine the number of terms in this sequence that will be exactly divisible by 5.

    Answer: The number of terms divisible by 5 is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Determine the last term in the sequence divisible by 5
    [−0 points ⇒ 4 / 4 points left]

    The first term in the sequence is 5, which is divisible by 5.

    But the last term in the sequence is 4,991, which is not divisible by 5. So we need to work out the value of the last term which is divisible by 5.

    Since the common difference of the sequence is 4, we can count back from 4,991 in steps of 4 until we reach a number which is divisible by 5.

    TIP: Use your calculator if necessary to find the last value which is divisible by 5.
    4,9914,9874,9834,9794,975

    So the last term which is divisible by 5 is 4,975.


    STEP: Continue the sequence to identify terms divisible by 5
    [−0 points ⇒ 4 / 4 points left]

    The original sequence is:

    5,1,3,7,11,15,19,23,27,31,35,,4,975

    The numbers divisible by 5 form a new sequence: 5,15,35,,4,975

    The new pattern is an arithmetic sequence with a=5 and d=20.


    STEP: Use the formula for arithmetic sequences to calculate the number of terms divisible by 5
    [−4 points ⇒ 0 / 4 points left]
    Tn=a+(n1)d4,975=5+(n1)(20)4,975=520n+205,000=20n250=n

    Submit your answer as:

ID is: 3858 Seed is: 6180

Finite arithmetic sequence

Adapted from DBE Nov 2016 Grade 12, P1, Q2
Maths formulas

Given the finite arithmetic sequence:

6,2,2,,90,94
  1. Write down the fourth term (T4) of the sequence.

    Answer: T4=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Determine the value of the fourth term
    [−1 point ⇒ 0 / 1 points left]

    We can see from the first three terms that the common difference value for this sequence is 4. So the fourth term can be calculated as:

    Tn=a+(n1)dT4=(6)+((4)1)(4)=6

    Submit your answer as:
  2. Calculate the number of terms in the sequence.

    Answer: n=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Calculate the number of terms in the sequence
    [−3 points ⇒ 0 / 3 points left]

    We can use the same formula that we used in Question 1 to find the number of terms in the sequence, because we know the value of the final term in the sequence:

    Tn=a+(n1)d94=(6)+(n1)(4)94=6+4n4104=4nn=26

    Submit your answer as:
  3. Calculate the sum of all the positive numbers in the sequence.

    Answer: The sum of the positive numbers is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Calculate the sum of the positive terms
    [−3 points ⇒ 0 / 3 points left]

    The first two terms are negative, and then all of the rest of the terms are positive. So we need to exclude the first two terms when we determine the sum of the positive terms.

    So the positive terms of the sequence look like:

    2,6,,94

    There will be 24 terms in this sequence, since we are excluding two negative terms. And the new value of a will be 2. We can then calculate the sum of these terms using the formula for the sum of an arithmetic series:

    Sn=n2[2a+(n1)d]S24=242[2(2)+((24)1)(4)]=48+1,152.048.0=1152

    Submit your answer as:
  4. Consider the sequence

    6,2,2,,90,94,,2,390

    Determine the number of terms in this sequence that will be exactly divisible by 3.

    Answer: The number of terms divisible by 3 is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Determine the last term in the sequence divisible by 3
    [−0 points ⇒ 4 / 4 points left]

    The first term in the sequence is 6, which is divisible by 3.

    But the last term in the sequence is 2,390, which is not divisible by 3. So we need to work out the value of the last term which is divisible by 3.

    Since the common difference of the sequence is 4, we can count back from 2,390 in steps of 4 until we reach a number which is divisible by 3.

    TIP: Use your calculator if necessary to find the last value which is divisible by 3.
    2,3902,3862,382

    So the last term which is divisible by 3 is 2,382.


    STEP: Continue the sequence to identify terms divisible by 3
    [−0 points ⇒ 4 / 4 points left]

    The original sequence is:

    6,2,2,6,10,14,18,,2,382

    The numbers divisible by 3 form a new sequence: 6,6,18,,2,382

    The new pattern is an arithmetic sequence with a=6 and d=12.


    STEP: Use the formula for arithmetic sequences to calculate the number of terms divisible by 3
    [−4 points ⇒ 0 / 4 points left]
    Tn=a+(n1)d2,382=6+(n1)(12)2,382=6+12n122,400=12n200=n

    Submit your answer as:

ID is: 2947 Seed is: 7848

Evaluating expressions involving arithmetic series

Evaluate without using a calculator

(876+28)+(16+15+14++4)
Answer:The expression is equal to:
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The question states that you should not use a calculator. That is a big hint that the expression can be simplified somehow so that the question becomes much easier than it looks.


STEP: Examine the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

The expression in the question is complex. However, the question says we must find the answer "without using a calculator". This means two things: (a) we must show all of the steps to get full marks and (b) there is probably a way to simplify the expression without doing lots of difficult calculations.

The first thing you hopefully noticed is that the expression is the sum of two different arithmetic series. The plan will be to use the formula Sn=n2(a+l) to rewrite both of the series: that will make the expression simpler. Then we will see what type of calculations we must do.


STEP: Find the number of terms in each series
[−2 points ⇒ 3 / 5 points left]

In order to use the formula Sn=n2(a+l), we need the values for n, a and l for each of the series. For both series, a and l are sitting in the question. However, the number of terms, n, is not obvious. We need to calculate the number of terms for both series. For this we can use the last term in each sequence in the formula Tn=a+(n1)d.

We must find the common difference (d) for each of the series first:

dA=7(8)=1dB=1516=1

Now find the number of terms in each series:

For 876+28:Tn=a+(n1)d28=8+(nA1)(1)36=(nA1)(1)36=nA137=nAFor 16+15+14++4:Tn=a+(n1)d4=16+(nB1)(1)12=(nB1)(1)12=nB113=nB

STEP: Use the sum formula Sn=n2(a+l) to rewrite the expression in the question
[−1 point ⇒ 2 / 5 points left]

Now use these values, together with a and l for each series, to rewrite the sum from the question using the formula, Sn=n2(a+l).

SA+SB=(876+28)+(16+15+14++4)=(372(8+28))+(132(16+4))

STEP: Simplify the expression (remember: no calculator!)
[−2 points ⇒ 0 / 5 points left]

From this point, we should start to simplify the expression. However, remember that the question says that we should evaluate the expression without a calculator. That means that somewhere during the calculations, there will probably be an opportunity to simplify the expression in some clever way which makes the calculation much easier... so keep your eyes open!

SA+SB=(372(8+28))+(132(16+4))=372(20)+132(20) factorise the 20's out! Check this: we can=(20)(372+132)=(20)(37+132)=(20)(502)=(20)(25)=500

By using factorisation, we were able to work out the question without the more complex calculations.

The final answer is 500.


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ID is: 2947 Seed is: 2870

Evaluating expressions involving arithmetic series

Evaluate the following quotient without using a calculator:

4+1244908886+10
Answer:The expression is equal to:
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The question states that you should not use a calculator. That is a big hint that the expression can be simplified somehow so that the question becomes much easier than it looks.


STEP: Examine the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

The expression in the question is complex. However, the question says we must find the answer "without using a calculator". This means two things: (a) we must show all of the steps to get full marks and (b) there is probably a way to simplify the expression without doing lots of difficult calculations.

The first thing you hopefully noticed is that the expression is the quotient of two different arithmetic series. The plan will be to use the formula Sn=n2(a+l) to rewrite both of the series: that will make the expression simpler. Then we will see what type of calculations we must do.


STEP: Find the number of terms in each series
[−2 points ⇒ 3 / 5 points left]

In order to use the formula Sn=n2(a+l), we need the values for n, a and l for each of the series. For both series, a and l are sitting in the question. However, the number of terms, n, is not obvious. We need to calculate the number of terms for both series. For this we can use the last term in each sequence in the formula Tn=a+(n1)d.

We must find the common difference (d) for each of the series first:

dA=14=3dB=88(90)=2

Now find the number of terms in each series:

For 4+1244:Tn=a+(n1)d44=4+(nA1)(3)48=(nA1)(3)16=nA117=nAFor 908886+10:Tn=a+(n1)d10=90+(nB1)(2)100=(nB1)(2)50=nB151=nB

STEP: Use the sum formula Sn=n2(a+l) to rewrite the expression in the question
[−1 point ⇒ 2 / 5 points left]

Now use these values, together with a and l for each series, to rewrite the quotient from the question using the formula, Sn=n2(a+l).

SASB=4+1244908886+10=(172)(444)(512)(90+10)

STEP: Simplify the expression (remember: no calculator!)
[−2 points ⇒ 0 / 5 points left]

From this point, we should start to simplify the expression. However, remember that the question says that we should evaluate the expression without a calculator. That means that somewhere during the calculations, there will probably be an opportunity to simplify the expression in some clever way which makes the calculation much easier... so keep your eyes open!

SASB=(172)(444)(512)(90+10)=(172)(40)(512)(80) cancel a factor of -40! Aha! We can=(172)(512)(12)=172(251)(12)=34204=16

Cool - we got the answer without having to do the more complicated calculations because there was an opportunity to cancel two large numbers (the -40's).

The final answer is 16.


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ID is: 2947 Seed is: 5487

Evaluating expressions involving arithmetic series

Evaluate the following sum without using a calculator:

(9101141)+(13172137)
Answer:The expression is equal to:
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The question states that you should not use a calculator. That is a big hint that the expression can be simplified somehow so that the question becomes much easier than it looks.


STEP: Examine the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

The expression in the question is complex. However, the question says we must find the answer "without using a calculator". This means two things: (a) we must show all of the steps to get full marks and (b) there is probably a way to simplify the expression without doing lots of difficult calculations.

The first thing you hopefully noticed is that the expression is the sum of two different arithmetic series. The plan will be to use the formula Sn=n2(a+l) to rewrite both of the series: that will make the expression simpler. Then we will see what type of calculations we must do.


STEP: Find the number of terms in each series
[−2 points ⇒ 3 / 5 points left]

In order to use the formula Sn=n2(a+l), we need the values for n, a and l for each of the series. For both series, a and l are sitting in the question. However, the number of terms, n, is not obvious. We need to calculate the number of terms for both series. For this we can use the last term in each sequence in the formula Tn=a+(n1)d.

We must find the common difference (d) for each of the series first:

dA=10(9)=1dB=17(13)=4

Now find the number of terms in each series:

For 9101141:Tn=a+(n1)d41=9+(nA1)(1)32=(nA1)(1)32=nA133=nAFor 13172137:Tn=a+(n1)d37=13+(nB1)(4)24=(nB1)(4)6=nB17=nB

STEP: Use the sum formula Sn=n2(a+l) to rewrite the expression in the question
[−1 point ⇒ 2 / 5 points left]

Now use these values, together with a and l for each series, to rewrite the sum from the question using the formula, Sn=n2(a+l).

SA+SB=(9101141)+(13172137)=(332(941))+(72(1337))

STEP: Simplify the expression (remember: no calculator!)
[−2 points ⇒ 0 / 5 points left]

From this point, we should start to simplify the expression. However, remember that the question says that we should evaluate the expression without a calculator. That means that somewhere during the calculations, there will probably be an opportunity to simplify the expression in some clever way which makes the calculation much easier... so keep your eyes open!

SA+SB=(332(941))+(72(1337))=332(50)+72(50) factorise the -50's out! Check this: we can=(50)(332+72)=(50)(33+72)=(50)(402)=(50)(20)=1,000

By using factorisation, we were able to work out the question without the more complex calculations.

The final answer is 1,000.


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ID is: 2938 Seed is: 1303

Using terms of an arithmetic sequence to find the sum

The fourth term of an arithmetic sequence is -5 and the sixth term is -15. Determine the sum of the first 49 terms of the sequence.

Answer:S49=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by using the two terms given in the question to find the common difference and then find the first term. After you have those numbers you can use one of the formulae for the sum of an arithmetic series.


STEP: Think about the information given in the question and consider how to solve the problem
[−0 points ⇒ 6 / 6 points left]

We have two terms in an arithmetic sequence, T4 and T6, and we must find the sum of 49 terms for the sequence. There are two formulae for the sum of an arithmetic sequence: Sn=n2(a+l) and Sn=n2(2a+(n1)d). You can see that we either need to find a and l to use the first version of the formula, or a and d to use the second version. Since the question does not tell us what the last term is, we will plan to use the second formula.


STEP: Use the two terms from the question to find the common difference of the sequence
[−3 points ⇒ 3 / 6 points left]

Think about how the two terms given fit into the entire sum:

T1+T2+T3+T4+T5+T6+
T1+T2+T3+(5)+T5+(15)+

This list can help you see that if we have any two terms from an arithmetic sequence, we can always use them to find the common difference, d, (because each term is always separated by d). Specifically, the difference between any two terms is related to the common difference.

Let us find the difference between T4 and T6: it is T6T4=(15)(5)=10. Now we use the formula Tn=a+(n1)d to do something clever:

T6T4=10(a+(61)d)(a+(41)d)=10for the terms we haveUse the values of n(a+5d)(a+3d)=10a+5da3d=102d=10d=102d=5

STEP: Use the common difference and the formula Tn=a+(n1)d to find the first term
[−1 point ⇒ 2 / 6 points left]

Now that we know the value of d, we can find the value of a. Do this using the formula Tn=a+(n1)d with either of the terms given in the question. We will use the first term, T4=5.

Tn=a+(n1)dT4=a+(41)d5=a+(3)(5)5=a1510=a

STEP: Calculate the sum using Sn=n2(2a+(n1)d)
[−2 points ⇒ 0 / 6 points left]

Finally we have enough information to evaluate the sum the question asked for. As noted at the beginning, we will use the formula Sn=n2(2a+(n1)d).

Substitute in the values and evaluate:

Sn=n2(2a+(n1)d)S49=492(2(10)+(491)(5))S49=492(20+(48)(5))S49=492(220)S49=5390

The sum of the first 49 terms is -5390.


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ID is: 2938 Seed is: 9959

Using terms of an arithmetic sequence to find the sum

Consider two terms of an arithmetic sequence: the sixth and eighth terms are 39 and 49, respectively. What is the sum of the first 21 terms of the sequence?

Answer:S21=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by using the two terms given in the question to find the common difference and then find the first term. After you have those numbers you can use one of the formulae for the sum of an arithmetic series.


STEP: Think about the information given in the question and consider how to solve the problem
[−0 points ⇒ 6 / 6 points left]

We have two terms in an arithmetic sequence, T6 and T8, and we must find the sum of 21 terms for the sequence. There are two formulae for the sum of an arithmetic sequence: Sn=n2(a+l) and Sn=n2(2a+(n1)d). You can see that we either need to find a and l to use the first version of the formula, or a and d to use the second version. Since the question does not tell us what the last term is, we will plan to use the second formula.


STEP: Use the two terms from the question to find the common difference of the sequence
[−3 points ⇒ 3 / 6 points left]

Think about how the two terms given fit into the entire sum:

+T5+T6+T7+T8+
+T5+39+T7+49+

This list can help you see that if we have any two terms from an arithmetic sequence, we can always use them to find the common difference, d, (because each term is always separated by d). Specifically, the difference between any two terms is related to the common difference.

Let us find the difference between T6 and T8: it is T8T6=4939=10. Now we use the formula Tn=a+(n1)d to do something clever:

T8T6=10(a+(81)d)(a+(61)d)=10for the terms we haveUse the values of n(a+7d)(a+5d)=10a+7da5d=102d=10d=102d=5

STEP: Use the common difference and the formula Tn=a+(n1)d to find the first term
[−1 point ⇒ 2 / 6 points left]

Now that we know the value of d, we can find the value of a. Do this using the formula Tn=a+(n1)d with either of the terms given in the question. We will use the first term, T6=39.

Tn=a+(n1)dT6=a+(61)d39=a+(5)539=a+2514=a

STEP: Calculate the sum using Sn=n2(2a+(n1)d)
[−2 points ⇒ 0 / 6 points left]

Finally we have enough information to evaluate the sum the question asked for. As noted at the beginning, we will use the formula Sn=n2(2a+(n1)d).

Substitute in the values and evaluate:

Sn=n2(2a+(n1)d)S21=212(2(14)+(211)5)S21=212(28+(20)5)S21=212(128)S21=1344

The sum of the first 21 terms is 1344.


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ID is: 2938 Seed is: 2088

Using terms of an arithmetic sequence to find the sum

Consider two terms of an arithmetic sequence: the fifth and eighth terms are 66 and 93, respectively. Determine the sum of the first 28 terms of the sequence.

Answer:S28=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by using the two terms given in the question to find the common difference and then find the first term. After you have those numbers you can use one of the formulae for the sum of an arithmetic series.


STEP: Think about the information given in the question and consider how to solve the problem
[−0 points ⇒ 6 / 6 points left]

We have two terms in an arithmetic sequence, T5 and T8, and we must find the sum of 28 terms for the sequence. There are two formulae for the sum of an arithmetic sequence: Sn=n2(a+l) and Sn=n2(2a+(n1)d). You can see that we either need to find a and l to use the first version of the formula, or a and d to use the second version. Since the question does not tell us what the last term is, we will plan to use the second formula.


STEP: Use the two terms from the question to find the common difference of the sequence
[−3 points ⇒ 3 / 6 points left]

Think about how the two terms given fit into the entire sum:

+T4+T5+T6+T7+T8+
+T4+66+T6+T7+93+

This list can help you see that if we have any two terms from an arithmetic sequence, we can always use them to find the common difference, d, (because each term is always separated by d). Specifically, the difference between any two terms is related to the common difference.

Let us find the difference between T5 and T8: it is T8T5=9366=27. Now we use the formula Tn=a+(n1)d to do something clever:

T8T5=27(a+(81)d)(a+(51)d)=27for the terms we haveUse the values of n(a+7d)(a+4d)=27a+7da4d=273d=27d=273d=9

STEP: Use the common difference and the formula Tn=a+(n1)d to find the first term
[−1 point ⇒ 2 / 6 points left]

Now that we know the value of d, we can find the value of a. Do this using the formula Tn=a+(n1)d with either of the terms given in the question. We will use the first term, T5=66.

Tn=a+(n1)dT5=a+(51)d66=a+(4)966=a+3630=a

STEP: Calculate the sum using Sn=n2(2a+(n1)d)
[−2 points ⇒ 0 / 6 points left]

Finally we have enough information to evaluate the sum the question asked for. As noted at the beginning, we will use the formula Sn=n2(2a+(n1)d).

Substitute in the values and evaluate:

Sn=n2(2a+(n1)d)S28=282(2(30)+(281)9)S28=14(60+(27)9)S28=14(303)S28=4242

The sum of the first 28 terms is 4242.


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ID is: 2953 Seed is: 7887

Finding the number of terms in a series

  1. The following expression shows an arithmetic series.

    5+13327331

    Determine the number of terms in the series.

    Answer: The number of terms is:
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by finding the values of a and d. Then use the formula Tn=a+(n1)d.


    STEP: Set up the equation using the last term
    [−1 point ⇒ 2 / 3 points left]

    The question gives us an arithmetic series and we must determine the number of terms in the series. We can do this using the formula Tn=a+(n1)d.

    The number of terms in the series is represented by the n-value of the last term in the series. In this case, the last term is 331. We can then substitute into the formula as follows:

    331=a+(nlast1)d

    If we also substitute in the values of the first term, a, and the common difference, d, we can solve the equation for the value of n for the last term. This will give us the total number of terms in the series.


    STEP: Find the values of a and d and substitute into the equation
    [−1 point ⇒ 1 / 3 points left]

    From the series in the question, we can get both the first term and the common difference. The first term is sitting there for us to take: a=5. The common difference requires a bit of work: it is the difference between any two consecutive terms in the series: d=T2T1, so we get d=15=4.

    Substitute these values into the equation from above:

    331=5+(nlast1)(4)

    STEP: Solve the equation for n
    [−1 point ⇒ 0 / 3 points left]

    Now we can solve the equation for nlast. Start by distributing the 4 into the brackets.

    331=5+(nlast1)(4)331=54nlast+4331=94nlast4nlast=340(nlast positiverearrange to make)nlast=85

    The n-value of the final term is 85.

    The calculation here shows that there are 85 terms in the series.


    Submit your answer as:
  2. Determine if 197 is one of the terms in the series or not. Use a calculation to justify your answer.

    NOTE: You will pick the correct choice below, but on a test or exam, you must show the calculation to get full marks.
    Answer:

    Is 197 in the series?

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can use the same method as in the first question to find the answer: substitute values in the equation Tn=a+(n1)d and find the value of n. The answer you get for n will tell you if 197 belongs in the series or not.


    STEP: Use the formula Tn=a+(n1)d to find the value of n
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to figure out whether or not 197 is a number in the series. In other words, if we write out every term in the series 5+13327331, will we find 197 somewhere in the middle or will the numbers "jump" over it?

    We can work out the answer using the same method as the first question - by finding the value of n for the number 197. Basically, we will assume that 197 is a term in the series and then use the value of n to check whether or not it is true! Substitute the values into the equation and calculate n.

    197=5+(n1)(4)197=54nlast+4197=94n4n=206(n positiverearrange to make)n=51.5

    This calculation shows that if the term is 197 then n=51.5.

    How does this calculation show us whether 197 is a term in the series or not? It depends on whether or not the value for n makes sense. In this case, we found that n51.5, which does not make sense because it is a decimal number (not a whole number). n represents the position of the term in the series. Finding a term in position n51.5 is the same as saying that you stood in a queue of 85 people and you were 51.5th person in the queue. That is impossible!

    The correct answer is no, the number is not a term in the series because the n-value is not a natural number.


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ID is: 2953 Seed is: 9482

Finding the number of terms in a series

  1. The following expression shows an arithmetic series.

    240+237+234++03

    How many terms are there in the series?

    Answer: The number of terms is:
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by finding the values of a and d. Then use the formula Tn=a+(n1)d.


    STEP: Set up the equation using the last term
    [−1 point ⇒ 2 / 3 points left]

    The question gives us an arithmetic series and we must determine the number of terms in the series. We can do this using the formula Tn=a+(n1)d.

    The number of terms in the series is represented by the n-value of the last term in the series. In this case, the last term is 3. We can then substitute into the formula as follows:

    3=a+(nlast1)d

    If we also substitute in the values of the first term, a, and the common difference, d, we can solve the equation for the value of n for the last term. This will give us the total number of terms in the series.


    STEP: Find the values of a and d and substitute into the equation
    [−1 point ⇒ 1 / 3 points left]

    From the series in the question, we can get both the first term and the common difference. The first term is sitting there for us to take: a=240. The common difference requires a bit of work: it is the difference between any two consecutive terms in the series: d=T2T1, so we get d=237240=3.

    Substitute these values into the equation from above:

    3=240+(nlast1)(3)

    STEP: Solve the equation for n
    [−1 point ⇒ 0 / 3 points left]

    Now we can solve the equation for nlast. Start by distributing the 3 into the brackets.

    3=240+(nlast1)(3)3=2403nlast+33=2433nlast3nlast=246(nlast positiverearrange to make)nlast=82

    The n-value of the final term is 82.

    The calculation here shows that there are 82 terms in the series.


    Submit your answer as:
  2. Is the number 135 a term in the series. Use a calculation to justify your answer.

    NOTE: You will pick the correct choice below, but on a test or exam, you must show the calculation to get full marks.
    Answer:

    Is 135 in the series?

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can use the same method as in the first question to find the answer: substitute values in the equation Tn=a+(n1)d and find the value of n. The answer you get for n will tell you if 135 belongs in the series or not.


    STEP: Use the formula Tn=a+(n1)d to find the value of n
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to figure out whether or not 135 is a number in the series. In other words, if we write out every term in the series 240+237+234++03, will we find 135 somewhere in the middle or will the numbers "jump" over it?

    We can work out the answer using the same method as the first question - by finding the value of n for the number 135. Basically, we will assume that 135 is a term in the series and then use the value of n to check whether or not it is true! Substitute the values into the equation and calculate n.

    135=240+(n1)(3)135=2403nlast+3135=2433n3n=108(n positiverearrange to make)n=36

    This calculation shows that if the term is 135 then n=36.

    How does this calculation show us whether 135 is a term in the series or not? It depends on whether or not the value for n makes sense. The answer n=36 does make sense because it is a natural number between 1 and 82. n=36 means that 135 is the 36th term in the series. (For comparison, suppose we got the answer n=36 or n=83; those answers are impossible, because we already know that the series starts at the first term, n=1, and ends at n=82.)

    The correct answer is yes, the number is a term in the series because the n-value is a natural number.


    Submit your answer as:

ID is: 2953 Seed is: 4553

Finding the number of terms in a series

  1. The following expression shows an arithmetic series.

    1+0+1++41+42

    Determine the number of terms in the series.

    Answer: The number of terms is:
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by finding the values of a and d. Then use the formula Tn=a+(n1)d.


    STEP: Set up the equation using the last term
    [−1 point ⇒ 2 / 3 points left]

    The question gives us an arithmetic series and we must determine the number of terms in the series. We can do this using the formula Tn=a+(n1)d.

    The number of terms in the series is represented by the n-value of the last term in the series. In this case, the last term is 42. We can then substitute into the formula as follows:

    42=a+(nlast1)d

    If we also substitute in the values of the first term, a, and the common difference, d, we can solve the equation for the value of n for the last term. This will give us the total number of terms in the series.


    STEP: Find the values of a and d and substitute into the equation
    [−1 point ⇒ 1 / 3 points left]

    From the series in the question, we can get both the first term and the common difference. The first term is sitting there for us to take: a=1. The common difference requires a bit of work: it is the difference between any two consecutive terms in the series: d=T2T1, so we get d=0(1)=1.

    Substitute these values into the equation from above:

    42=1+(nlast1)(1)

    STEP: Solve the equation for n
    [−1 point ⇒ 0 / 3 points left]

    Now we can solve the equation for nlast. Start by distributing the 1 into the brackets.

    42=1+(nlast1)(1)42=1+nlast142=2+nlast44=nlast

    The n-value of the final term is 44.

    The calculation here shows that there are 44 terms in the series.


    Submit your answer as:
  2. Determine if 37 is one of the terms in the series or not. Use a calculation to justify your answer.

    NOTE: You will pick the correct choice below, but on a test or exam, you must show the calculation to get full marks.
    Answer:

    Is 37 in the series?

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can use the same method as in the first question to find the answer: substitute values in the equation Tn=a+(n1)d and find the value of n. The answer you get for n will tell you if 37 belongs in the series or not.


    STEP: Use the formula Tn=a+(n1)d to find the value of n
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to figure out whether or not 37 is a number in the series. In other words, if we write out every term in the series 1+0+1++41+42, will we find 37 somewhere in the middle or will the numbers "jump" over it?

    For the series 1+0+1++41+42, it is clear that the number 37 is in the series. However, we will show that it is true with a calculation, as the question instructs.

    We can work out the answer using the same method as the first question - by finding the value of n for the number 37. Basically, we will assume that 37 is a term in the series and then use the value of n to check whether or not it is true! Substitute the values into the equation and calculate n.

    37=1+(n1)(1)37=1+nlast137=2+n39=n

    This calculation shows that if the term is 37 then n=39.

    How does this calculation show us whether 37 is a term in the series or not? It depends on whether or not the value for n makes sense. The answer n=39 does make sense because it is a natural number between 1 and 44. n=39 means that 37 is the 39th term in the series. (For comparison, suppose we got the answer n=39 or n=45; those answers are impossible, because we already know that the series starts at the first term, n=1, and ends at n=44.)

    The correct answer is yes, the number is a term in the series because the n-value is a natural number.


    Submit your answer as:

4. Geometric sequences


ID is: 3923 Seed is: 7923

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.2
Maths formulas

The following geometric sequence is given:

14;7;3.5;1.75;

Which of the following is the correct expression for the nth term of this sequence?

A Tn=14(12)n
B Tn=14(12)n1
C Tn=14(7)n1
D Tn=14+(12)(n1)
Answer: The correct choice is Option .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

First determine the values of the first term and the constant ratio of the sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

We need to know the values of a and r in order to identify the correct equation.

a is the first term of the sequence, so here a=14.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=714=12

STEP: Substitute a and r into the general formula
[−1 point ⇒ 0 / 2 points left]

Substitute these values into the formula for the general term of a geometric sequence, so we can identify the correct expression:

Tn=arn1=14(12)n1

So the correct choice is Option B.


Submit your answer as:

ID is: 3923 Seed is: 2363

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.2
Maths formulas

The following geometric sequence is given:

12;6;3;1.5;

Which of the following is the correct expression for the nth term of this sequence?

A Tn=12(2)n
B Tn=12+(12)(n1)
C Tn=12(12)n1
D Tn=12(12)n1
Answer: The correct choice is Option .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

First determine the values of the first term and the constant ratio of the sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

We need to know the values of a and r in order to identify the correct equation.

a is the first term of the sequence, so here a=12.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=612=12

STEP: Substitute a and r into the general formula
[−1 point ⇒ 0 / 2 points left]

Substitute these values into the formula for the general term of a geometric sequence, so we can identify the correct expression:

Tn=arn1=12(12)n1

So the correct choice is Option D.


Submit your answer as:

ID is: 3923 Seed is: 7905

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.2
Maths formulas

The following geometric sequence is given:

16;8;4;2;

Which of the following is the correct expression for the nth term of this sequence?

A Tn=16+(12)(n1)
B Tn=16(12)n
C Tn=16(12)n1
D Tn=16(8)n1
Answer: The correct choice is Option .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

First determine the values of the first term and the constant ratio of the sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

We need to know the values of a and r in order to identify the correct equation.

a is the first term of the sequence, so here a=16.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=816=12

STEP: Substitute a and r into the general formula
[−1 point ⇒ 0 / 2 points left]

Substitute these values into the formula for the general term of a geometric sequence, so we can identify the correct expression:

Tn=arn1=16(12)n1

So the correct choice is Option C.


Submit your answer as:

ID is: 2954 Seed is: 1491

Constant ratio and general formula

  1. The first three terms of a geometric sequence are given:

    12;1;2;
    1. Calculate the constant ratio, r.
    2. Calculate the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    numeric
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The constant ratio, r, is the quotient of one term and the term before it. For example,

    r=T3T2

    STEP: Calculate the value of r
    [−1 point ⇒ 1 / 2 points left]

    We have the geometric sequence 12;1;2; and we need to figure out the constant ratio as well as the general formula. The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can calculate the ratio by finding the quotient of any two consecutive terms: r=T3T2 or T2T1 or TnTn1. Therefore, we can pick any pair of consecutive terms from the sequence. This time let's use the second and third terms:

    r=T3T2r=21r=2

    Notice that the ratio is negative: that is the reason why the terms in the sequence keep changing signs. (Multiplying by a negative always changes the sign.)


    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 2 points left]

    Now that we have calculated the value for r, we can calculate the general formula for 12;1;2;. To find the general formula for a geometric sequence, we use the equation Tn=arn1. The variable a is the first term in the sequence and r is the constant ratio.

    The question states that the first three terms of the sequence are 12;1;2. So, the value for a is 12. Substituting a=12 and r=2 into the general formula, we get:

    Tn=arn1=(12)(2)n1
    NOTE:

    While it is not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Remember that we calculate exponents before multiplying (according to BODMAS) so do not multiply 12 and 2. Start by simplifying the exponential part of the expression, (2)n1.

    Tn=(12)(2)n1=(12)(2)n(2)1=(12)(2)n(12)=14(2)n

    Therefore for the given sequence, the general formula can be written as Tn=14(2)n


    Submit your answer as: and
  2. The first three terms of a geometric sequence are given:

    3;3x;3x2;
    1. Calculate the constant ratio, r.
    2. Calculate the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of x and n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    expression
    one-of
    type(expression)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do not let the variables in the sequence bother you: this question can be solved in the same way as the first question.


    STEP: Calcuate the value of r
    [−2 points ⇒ 1 / 3 points left]

    As with the first question, we have a geometric sequence, 3;3x;3x2;, and we need to figure out the constant ratio as well as the general formula. We calculate the ratio for this sequence in the same way as for the previous question: use any two consecutive terms: r=T3T2 or T2T1 or Tn1Tn. Since we can choose which terms we use, pick the simplest ones available for the calculation. Then simplify.

    r=T3T2r=3x23xr=x

    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 3 points left]

    Now we will find the general formula for the sequence. This follows the exact same process as in the first question: we use Tn=arn1 and substitute in the correct values for a and r for this sequence. The first term of the sequence is a=3 while the constant ratio is r=x. Substitute these values into the formula:

    Tn=arn1=(3)(x)n1
    NOTE:

    While it is also not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Start by simplifying (x)n1 using the exponent law.

    Tn=(3)(x)n1=(3)(x)n(x)1=(3)(1x)(x)n=(3x)(x)n

    Therefore for the given sequence, the general formula can be written as Tn=(3x)(x)n.


    Submit your answer as: and

ID is: 2954 Seed is: 2172

Constant ratio and general formula

  1. The first three terms of a geometric sequence are given:

    7;21;63;
    1. Find the constant ratio, r.
    2. Determine the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    numeric
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The constant ratio, r, is the quotient of one term and the term before it. For example,

    r=T3T2

    STEP: Calculate the value of r
    [−1 point ⇒ 1 / 2 points left]

    We have the geometric sequence 7;21;63; and we need to figure out the constant ratio as well as the general formula. The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can calculate the ratio by finding the quotient of any two consecutive terms: r=T3T2 or T2T1 or TnTn1. Therefore, we can pick any pair of consecutive terms from the sequence. This time let's use the first and second terms:

    r=T2T1r=217=3

    Notice that the ratio is negative: that is the reason why the terms in the sequence keep changing signs. (Multiplying by a negative always changes the sign.)


    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 2 points left]

    Now that we have calculated the value for r, we can calculate the general formula for 7;21;63;. To find the general formula for a geometric sequence, we use the equation Tn=arn1. The variable a is the first term in the sequence and r is the constant ratio.

    The question states that the first three terms of the sequence are 7;21;63. So, the value for a is 7. Substituting a=7 and r=3 into the general formula, we get:

    Tn=arn1=(7)(3)n1
    NOTE:

    While it is not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Remember that we calculate exponents before multiplying (according to BODMAS) so do not multiply 7 and 3. Start by simplifying the exponential part of the expression, (3)n1.

    Tn=(7)(3)n1=(7)(3)n(3)1=(7)(3)n(13)=73(3)n

    Therefore for the given sequence, the general formula can be written as Tn=73(3)n


    Submit your answer as: and
  2. The first three terms of a geometric sequence are given:

    5;5x2;5x4;
    1. Find the constant ratio, r.
    2. Determine the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of x and n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    expression
    one-of
    type(expression)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do not let the variables in the sequence bother you: this question can be solved in the same way as the first question.


    STEP: Calcuate the value of r
    [−2 points ⇒ 1 / 3 points left]

    As with the first question, we have a geometric sequence, 5;5x2;5x4;, and we need to figure out the constant ratio as well as the general formula. We calculate the ratio for this sequence in the same way as for the previous question: use any two consecutive terms: r=T3T2 or T2T1 or Tn1Tn. Since we can choose which terms we use, pick the simplest ones available for the calculation. Then simplify.

    r=T2T1r=5x25=x2

    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 3 points left]

    Now we will find the general formula for the sequence. This follows the exact same process as in the first question: we use Tn=arn1 and substitute in the correct values for a and r for this sequence. The first term of the sequence is a=5 while the constant ratio is r=x2. Substitute these values into the formula:

    Tn=arn1=(5)(x2)n1
    NOTE:

    While it is also not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Start by simplifying (x2)n1 using the exponent law.

    Tn=(5)(x2)n1=(5)(x2)n(x2)1=(5)(1x2)(x2)n=(5x2)(x2)n

    Therefore for the given sequence, the general formula can be written as Tn=(5x2)(x2)n.


    Submit your answer as: and

ID is: 2954 Seed is: 6279

Constant ratio and general formula

  1. The first three terms of a geometric sequence are given:

    13;1;3;
    1. Find the constant ratio, r.
    2. Determine the general formula for the nth term of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    numeric
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The constant ratio, r, is the quotient of one term and the term before it. For example,

    r=T3T2

    STEP: Calculate the value of r
    [−1 point ⇒ 1 / 2 points left]

    We have the geometric sequence 13;1;3; and we need to figure out the constant ratio as well as the general formula. The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can calculate the ratio by finding the quotient of any two consecutive terms: r=T3T2 or T2T1 or TnTn1. Therefore, we can pick any pair of consecutive terms from the sequence. This time let's use the second and third terms:

    r=T3T2r=31r=3

    Notice that the ratio is negative: that is the reason why the terms in the sequence keep changing signs. (Multiplying by a negative always changes the sign.)


    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 2 points left]

    Now that we have calculated the value for r, we can calculate the general formula for 13;1;3;. To find the general formula for a geometric sequence, we use the equation Tn=arn1. The variable a is the first term in the sequence and r is the constant ratio.

    The question states that the first three terms of the sequence are 13;1;3. So, the value for a is 13. Substituting a=13 and r=3 into the general formula, we get:

    Tn=arn1=(13)(3)n1
    NOTE:

    While it is not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Remember that we calculate exponents before multiplying (according to BODMAS) so do not multiply 13 and 3. Start by simplifying the exponential part of the expression, (3)n1.

    Tn=(13)(3)n1=(13)(3)n(3)1=(13)(3)n(13)=19(3)n

    Therefore for the given sequence, the general formula can be written as Tn=19(3)n


    Submit your answer as: and
  2. The first three terms of a geometric sequence are given:

    4x;8x;16x;
    1. Find the constant ratio, r.
    2. Determine the general formula for the nth term of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of x and n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    expression
    one-of
    type(expression)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do not let the variables in the sequence bother you: this question can be solved in the same way as the first question.


    STEP: Calcuate the value of r
    [−2 points ⇒ 1 / 3 points left]

    As with the first question, we have a geometric sequence, 4x;8x;16x;, and we need to figure out the constant ratio as well as the general formula. We calculate the ratio for this sequence in the same way as for the previous question: use any two consecutive terms: r=T3T2 or T2T1 or Tn1Tn. Since we can choose which terms we use, pick the simplest ones available for the calculation. Then simplify.

    r=T2T1r=8x4x=2

    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 3 points left]

    Now we will find the general formula for the sequence. This follows the exact same process as in the first question: we use Tn=arn1 and substitute in the correct values for a and r for this sequence. The first term of the sequence is a=4x while the constant ratio is r=2. Substitute these values into the formula:

    Tn=arn1=(4x)(2)n1
    NOTE:

    While it is also not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Start by simplifying (2)n1 using the exponent law.

    Tn=(4x)(2)n1=(4x)(2)n(2)1=(4x)(12)(2)n=(2x)(2)n

    Therefore for the given sequence, the general formula can be written as Tn=(2x)(2)n.


    Submit your answer as: and

ID is: 2951 Seed is: 240

The geometric mean

  1. What is the geometric mean of the following numbers if the mean is negative?

    12 and 18

    Give all possible answers.

    INSTRUCTION: If there is more than one answer, separate the answers by the ';' symbol.
    Answer: The geometric mean is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    The geometric mean of 12 and 18 is a number y which makes the numbers 12;y;18 follow a geometric pattern. Use the constant ratio to write an equation which includes y. Then solve the equation.
    STEP: Write an equation based on the constant ratio of the terms
    [−1 point ⇒ 1 / 2 points left]

    The question asks us to find the geometric mean for the numbers 12 and 18.

    This question is the same as: 'Find the value of y so that the terms 12;y;18 follow a geometric sequence.' This is because if 12;y;18 form a geometric sequence, then y is the geometric mean of 12 and 18. (If you have three consecutive terms of a geometric sequence, the middle term is always the geometric mean of the first and third terms.)

    Since we are dealing with a geometric sequence, we can find the answer using the constant ratio. Remember that the constant ratio is the quotient of any two consecutive terms in the sequence. We do not know the value of the constant ratio, but we do know that r is equal to both T2T1 and T3T2. So we can write the equation below with the first three terms and then substitute in the terms in this sequence:

    T2T1=T3T2y(12)=(18)y

    The equation above summarizes the relationship between the terms based on the constant ratio.


    STEP: Find the final answer or answers
    [−1 point ⇒ 0 / 2 points left]

    Now we can solve the equation for y.

    y(12)=(18)yy2=(18)(12)y2=116y=±116which comes with the square root!Don't forget the plus-minusy=±14

    We get two answers because of the ± which comes into the solution with the square-root step. This makes sense, because it is possible for the terms of a geometric sequence to change signs: if the constant ratio is positive, the sequence will be 12;14;18 while if the constant ratio is negative the sequence will be 12;14;18.

    NOTE:

    There is a formula for the geometric mean of two numbers: the geometric means of a and b are x=±ab. For more about the connection between this formula and the working shown above, see the explanation in the textbook.

    There are two answers at this point, but the question stated that the mean is negative. So we can ignore the 14 answer.

    The correct answer is 14.


    Submit your answer as:
  2. Determine the common ratio for the three terms in Question 1.

    INSTRUCTION:If there is more than one answer, separate the answers by the ';' symbol.
    Answer: Common ratio =
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    The common ratio is the quotient of any two consecutive terms in the sequence.
    STEP: Use two consecutive terms to calculate r
    [−3 points ⇒ 0 / 3 points left]

    We need to find the common ratio for the geometric sequence. We know from Question 1 that the sequence is 12;14;18.

    The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can use any two consecutive terms in the sequence to find the answer:

    r=T3T2orr=T2T1

    We can summarise this relationship as r=TnTn1. In this solution we will use the second and third terms.

    r=T3T2r=(18)(14)r=12

    Now we can see why the terms in the sequence alternate signs: the constant ratio is negative.

    The constant ratio is r=12.


    Submit your answer as:

ID is: 2951 Seed is: 1428

The geometric mean

  1. What is the geometric mean of the following numbers if the mean is negative?

    9 and 225

    Give all possible answers.

    INSTRUCTION: If there is more than one answer, separate the answers by the ';' symbol.
    Answer: The geometric mean is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    The geometric mean of 9 and 225 is a number x which makes the numbers 9;x;225 follow a geometric pattern. Use the constant ratio to write an equation which includes x. Then solve the equation.
    STEP: Write an equation based on the constant ratio of the terms
    [−1 point ⇒ 1 / 2 points left]

    The question asks us to find the geometric mean for the numbers 9 and 225.

    This question is the same as: 'Find the value of x so that the terms 9;x;225 follow a geometric sequence.' This is because if 9;x;225 form a geometric sequence, then x is the geometric mean of 9 and 225. (If you have three consecutive terms of a geometric sequence, the middle term is always the geometric mean of the first and third terms.)

    Since we are dealing with a geometric sequence, we can find the answer using the constant ratio. Remember that the constant ratio is the quotient of any two successive terms in the sequence. We do not know the value of the constant ratio, but we do know that r is equal to both T2T1 and T3T2. So we can write the equation below with the first three terms and then substitute in the terms in this sequence:

    T2T1=T3T2x(9)=(225)x

    The equation above summarizes the relationship between the terms based on the constant ratio.


    STEP: Find the final answer or answers
    [−1 point ⇒ 0 / 2 points left]

    Now we can solve the equation for x.

    x(9)=(225)xx2=(225)(9)x2=2025x=±2025which comes with the square root!Don't forget the plus-minusx=±45

    We get two answers because of the ± which comes into the solution with the square-root step. This makes sense, because it is possible for the terms of a geometric sequence to change signs: if the constant ratio is positive, the sequence will be 9;45;225 while if the constant ratio is negative the sequence will be 9;45;225.

    NOTE:

    There is a formula for the geometric mean of two numbers: the geometric means of a and b are x=±ab. For more about the connection between this formula and the working shown above, see the explanation in the textbook.

    There are two answers at this point, but the question stated that the mean is negative. So we can ignore the 45 answer.

    The correct answer is 45.


    Submit your answer as:
  2. Find the general formula for the terms in the sequence in Question 1 if it is now given that the constant ratio of the sequence is 5.

    Answer: Tn=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    The answer will be an expression which includes the variable n.
    STEP: Use the formula Tn=arn1 to find the answer
    [−3 points ⇒ 0 / 3 points left]

    The question asks us to find the general formula for this sequence. The general formula is an expression which tells us how to calculate any term in the sequence.

    Start with the equation:

    Tn=arn1

    a represents the first term in the sequence, r represents the constant ratio, and n represents the rank (position) of the term in the sequence. For the sequence 9;45;225, the first term is a=9. The question tells us that the constant ratio is r=5. Substitute these numbers into the equation and simplify.

    Tn=(9)(5)n1=(9)(5)n(5)1=(9)(15)(5)n=(95)(5)n

    The general formula for the sequence is Tn=95(5)n.


    Submit your answer as:

ID is: 2951 Seed is: 3691

The geometric mean

  1. Find the geometric mean of these two numbers if the mean is negative:

    5 and 45

    Give all possible answers.

    INSTRUCTION: If there is more than one answer, separate the answers by the ';' symbol.
    Answer: The geometric mean is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    The geometric mean of 5 and 45 is a number x which makes the numbers 5;x;45 follow a geometric pattern. Use the constant ratio to write an equation which includes x. Then solve the equation.
    STEP: Write an equation based on the constant ratio of the terms
    [−1 point ⇒ 1 / 2 points left]

    The question asks us to find the geometric mean for the numbers 5 and 45.

    This question is the same as: 'Find the value of x so that the terms 5;x;45 follow a geometric sequence.' This is because if 5;x;45 form a geometric sequence, then x is the geometric mean of 5 and 45. (If you have three consecutive terms of a geometric sequence, the middle term is always the geometric mean of the first and third terms.)

    Since we are dealing with a geometric sequence, we can find the answer using the constant ratio. Remember that the constant ratio is the quotient of any two consecutive terms in the sequence. We do not know the value of the constant ratio, but we do know that r is equal to both T2T1 and T3T2. So we can write the equation below with the first three terms and then substitute in the terms in this sequence:

    T2T1=T3T2x5=45x

    The equation above summarizes the relationship between the terms based on the constant ratio.


    STEP: Find the final answer or answers
    [−1 point ⇒ 0 / 2 points left]

    Now we can solve the equation for x.

    x5=45xx2=(45)(5)x2=225x=±225which comes with the square root!Don't forget the plus-minusx=±15

    We get two answers because of the ± which comes into the solution with the square-root step. This makes sense, because it is possible for the terms of a geometric sequence to change signs: if the constant ratio is positive, the sequence will be 5;15;45 while if the constant ratio is negative the sequence will be 5;15;45.

    NOTE:

    There is a formula for the geometric mean of two numbers: the geometric means of a and b are x=±ab. For more about the connection between this formula and the working shown above, see the explanation in the textbook.

    There are two answers at this point, but the question stated that the mean is negative. So we can ignore the 15 answer.

    The correct answer is 15.


    Submit your answer as:
  2. Determine the common ratio for the three terms in Question 1.

    INSTRUCTION:If there is more than one answer, separate the answers by the ';' symbol.
    Answer: Common ratio =
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    The common ratio is the quotient of any two successive terms in the sequence.
    STEP: Use two consecutive terms to calculate r
    [−3 points ⇒ 0 / 3 points left]

    We need to find the common ratio for the geometric sequence. We know from Question 1 that the sequence is 5;15;45.

    The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can use any two consecutive terms in the sequence to find the answer:

    r=T3T2orr=T2T1

    We can summarise this relationship as r=TnTn1. In this solution we will use the second and third terms.

    r=T3T2r=45(15)r=3

    Now we can see why the terms in the sequence alternate signs: the constant ratio is negative.

    The constant ratio is r=3.


    Submit your answer as:

ID is: 3925 Seed is: 8387

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.1
Maths formulas

The following geometric sequence is given:

8;4;2;1;

Calculate the value of the sixth term, T6, of this sequence.

INSTRUCTION: Your answer should be exact - do not round off.
Answer: T6=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the general term of a geometric sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

Before we can use this formula to find the sixth term of the sequence, we need to know the values of a and r.

a is the first term of the sequence, so here a=8.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=48=12

STEP: Calculate the value of the sixth term
[−1 point ⇒ 0 / 2 points left]

We can now use these values in the formula for the general term, to find the value of the sixth term.

T6=(8)(12)(61)=(8)(132)=0.25

Submit your answer as:

ID is: 3925 Seed is: 8924

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.1
Maths formulas

The following geometric sequence is given:

32;8;2;0.5;

Calculate the value of the fifth term, T5, of this sequence.

INSTRUCTION: Your answer should be exact - do not round off.
Answer: T5=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the general term of a geometric sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

Before we can use this formula to find the fifth term of the sequence, we need to know the values of a and r.

a is the first term of the sequence, so here a=32.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=832=14

STEP: Calculate the value of the fifth term
[−1 point ⇒ 0 / 2 points left]

We can now use these values in the formula for the general term, to find the value of the fifth term.

T5=(32)(14)(51)=(32)(1256)=0.125

Submit your answer as:

ID is: 3925 Seed is: 4618

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.1
Maths formulas

The following geometric sequence is given:

20;10;5;2.5;

Calculate the value of the sixth term, T6, of this sequence.

INSTRUCTION: Your answer should be exact - do not round off.
Answer: T6=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the general term of a geometric sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

Before we can use this formula to find the sixth term of the sequence, we need to know the values of a and r.

a is the first term of the sequence, so here a=20.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=1020=12

STEP: Calculate the value of the sixth term
[−1 point ⇒ 0 / 2 points left]

We can now use these values in the formula for the general term, to find the value of the sixth term.

T6=(20)(12)(61)=(20)(132)=0.625

Submit your answer as:

5. Geometric series

6. Practical applications


ID is: 3871 Seed is: 5397

Stripes!

Adapted from DBE Nov 2016 Grade 12, P1, Q3.2
Maths formulas

Rectangles of width 1 cm are drawn from the edge of a sheet of paper that is 38 cm long, such that there is a 1 cm gap between one rectangle and the next. The length of the first rectangle is 21 cm, and the length of each successive rectangle is 90% of the length of the previous rectangle, until there are rectangles drawn along the entire length of AD. Each rectangle is coloured blue.

Answer the following questions about this scenario.

INSTRUCTION: Round all answers to two decimal places.
  1. Calculate the length of the 15th rectangle.

    Answer: The length is cm.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise geometric sequences in the Everything Maths textbook.


    STEP: Use the general formula for a geometric sequence
    [−3 points ⇒ 0 / 3 points left]

    Each rectangle is 90% smaller than the one before it. Since there is a constant ratio between the lengths of the rectangles, we can use the general formula for a geometric sequence

    Tn=arn1

    to find the length of the 15th rectangle. The length of the first rectangle is a=21 cm and the constant ratio is r=0.9, so we can substitute and solve for Tn:

    Tn=arn1=(21)(0.9)14=4.80412...4.8 cm

    Submit your answer as:
  2. Calculate the percentage of the paper that is coloured blue.

    Answer:

    The percentage of paper that is coloured blue is %.

    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite geometric series in the Everything Maths textbook.


    STEP: Calculate the total area of the blue rectangles
    [−2 points ⇒ 2 / 4 points left]

    The lengths of the blue rectangles form a geometric series. So we can use the formula for the sum of a geometric series to calculate the total length of the rectangles on the page.

    Sn=a(1rn)(1r)

    The only value that we need to determine before we can use this calculation is the total number of rectangles on the page, n. Since the page is 38 cm high, and each rectangle has a height of 1 cm, with a space of 1 cm between each rectangle, there must be a total of n=19 rectangles going up the page.

    Substituting into the formula, we can calculate the sum of the lengths of the rectangles:

    Sn=a(1rn)(1r)=(21)(1(0.9)19)(1(0.9))=(21)(0.86491...)(0.1)=181.63211...181.63 cm

    Since the height of each of the rectangles is 1 cm, the total area of the rectangles is easy to calculate.

    A=181.63 cm×1 cm=181.63 cm2

    STEP: Calculate the percentage of paper that is covered
    [−2 points ⇒ 0 / 4 points left]

    The total area of the page is

    Apage=38×21=798 cm2

    So the percentage of the page that is coloured blue is:

    blue=181.63798×100%=22.76065...=22.76%

    Submit your answer as:

ID is: 3871 Seed is: 2724

Stripes!

Adapted from DBE Nov 2016 Grade 12, P1, Q3.2
Maths formulas

Rectangles of width 1 cm are drawn from the edge of a sheet of paper that is 34 cm long, such that there is a 1 cm gap between one rectangle and the next. The length of the first rectangle is 23 cm, and the length of each successive rectangle is 80% of the length of the previous rectangle, until there are rectangles drawn along the entire length of AD. Each rectangle is coloured blue.

Answer the following questions about this scenario.

INSTRUCTION: Round all answers to two decimal places.
  1. Calculate the length of the 12th rectangle.

    Answer: The length is cm.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise geometric sequences in the Everything Maths textbook.


    STEP: Use the general formula for a geometric sequence
    [−3 points ⇒ 0 / 3 points left]

    Each rectangle is 80% smaller than the one before it. Since there is a constant ratio between the lengths of the rectangles, we can use the general formula for a geometric sequence

    Tn=arn1

    to find the length of the 12th rectangle. The length of the first rectangle is a=23 cm and the constant ratio is r=0.8, so we can substitute and solve for Tn:

    Tn=arn1=(23)(0.8)11=1.97568...1.98 cm

    Submit your answer as:
  2. Calculate the percentage of the paper that is coloured blue.

    Answer:

    The percentage of paper that is coloured blue is %.

    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite geometric series in the Everything Maths textbook.


    STEP: Calculate the total area of the blue rectangles
    [−2 points ⇒ 2 / 4 points left]

    The lengths of the blue rectangles form a geometric series. So we can use the formula for the sum of a geometric series to calculate the total length of the rectangles on the page.

    Sn=a(1rn)(1r)

    The only value that we need to determine before we can use this calculation is the total number of rectangles on the page, n. Since the page is 34 cm high, and each rectangle has a height of 1 cm, with a space of 1 cm between each rectangle, there must be a total of n=17 rectangles going up the page.

    Substituting into the formula, we can calculate the sum of the lengths of the rectangles:

    Sn=a(1rn)(1r)=(23)(1(0.8)17)(1(0.8))=(23)(0.97748...)(0.2)=112.41043...112.41 cm

    Since the height of each of the rectangles is 1 cm, the total area of the rectangles is easy to calculate.

    A=112.41 cm×1 cm=112.41 cm2

    STEP: Calculate the percentage of paper that is covered
    [−2 points ⇒ 0 / 4 points left]

    The total area of the page is

    Apage=34×23=782 cm2

    So the percentage of the page that is coloured blue is:

    blue=112.41782×100%=14.37468...=14.37%

    Submit your answer as:

ID is: 3871 Seed is: 4902

Stripes!

Adapted from DBE Nov 2016 Grade 12, P1, Q3.2
Maths formulas

Rectangles of width 1 cm are drawn from the edge of a sheet of paper that is 20 cm long, such that there is a 1 cm gap between one rectangle and the next. The length of the first rectangle is 22 cm, and the length of each successive rectangle is 70% of the length of the previous rectangle, until there are rectangles drawn along the entire length of AD. Each rectangle is coloured blue.

Answer the following questions about this scenario.

INSTRUCTION: Round all answers to two decimal places.
  1. Calculate the length of the seventh rectangle.

    Answer: The length is cm.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise geometric sequences in the Everything Maths textbook.


    STEP: Use the general formula for a geometric sequence
    [−3 points ⇒ 0 / 3 points left]

    Each rectangle is 70% smaller than the one before it. Since there is a constant ratio between the lengths of the rectangles, we can use the general formula for a geometric sequence

    Tn=arn1

    to find the length of the seventh rectangle. The length of the first rectangle is a=22 cm and the constant ratio is r=0.7, so we can substitute and solve for Tn:

    Tn=arn1=(22)(0.7)6=2.58827...2.59 cm

    Submit your answer as:
  2. Calculate the percentage of the paper that is coloured blue.

    Answer:

    The percentage of paper that is coloured blue is %.

    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite geometric series in the Everything Maths textbook.


    STEP: Calculate the total area of the blue rectangles
    [−2 points ⇒ 2 / 4 points left]

    The lengths of the blue rectangles form a geometric series. So we can use the formula for the sum of a geometric series to calculate the total length of the rectangles on the page.

    Sn=a(1rn)(1r)

    The only value that we need to determine before we can use this calculation is the total number of rectangles on the page, n. Since the page is 20 cm high, and each rectangle has a height of 1 cm, with a space of 1 cm between each rectangle, there must be a total of n=10 rectangles going up the page.

    Substituting into the formula, we can calculate the sum of the lengths of the rectangles:

    Sn=a(1rn)(1r)=(22)(1(0.7)10)(1(0.7))=(22)(0.97175...)(0.3)=71.26184...71.26 cm

    Since the height of each of the rectangles is 1 cm, the total area of the rectangles is easy to calculate.

    A=71.26 cm×1 cm=71.26 cm2

    STEP: Calculate the percentage of paper that is covered
    [−2 points ⇒ 0 / 4 points left]

    The total area of the page is

    Apage=20×22=440 cm2

    So the percentage of the page that is coloured blue is:

    blue=71.26440×100%=16.19545...=16.2%

    Submit your answer as:

ID is: 2942 Seed is: 7874

Testing for convergent and divergent series

  1. Consider the series shown below and calculate r.

    8+32+128+512+
    Answer: r=
    numeric
    2 attempts remaining
    STEP: Determine the value of r
    [−2 points ⇒ 0 / 2 points left]

    To calculate the constant ratio (r), we need to determine how consecutive terms in the series are related to each other:

    r=T2T1=328=4Or r=T3T2=12832=4

    Therefore, the constant ratio for this geometric series is r=4 .


    Submit your answer as:
  2. Does the given series diverge or converge?

    Answer: The series .
    STEP: Use the test for convergence
    [−1 point ⇒ 0 / 1 points left]

    To test whether an infinte geometric series converges or diverges, we look at the value of the constant ratio:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=4, we can conclude that the series diverges.


    Submit your answer as:

ID is: 2942 Seed is: 9324

Testing for convergent and divergent series

  1. Given the following geometric series, determine the value of the constant ratio r.

    4+(8)+16+(32)+
    Answer: r=
    numeric
    2 attempts remaining
    STEP: Determine the value of r
    [−2 points ⇒ 0 / 2 points left]

    To calculate the constant ratio (r), we need to determine how consecutive terms in the series are related to each other:

    r=T2T1=84=2Or r=T3T2=168=2

    Therefore, the constant ratio for this geometric series is r=2 .


    Submit your answer as:
  2. Is this a divergent or convergent series?

    Answer: The series .
    STEP: Use the test for convergence
    [−1 point ⇒ 0 / 1 points left]

    To test whether an infinte geometric series converges or diverges, we look at the value of the constant ratio:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=2, we can conclude that the series diverges.


    Submit your answer as:

ID is: 2942 Seed is: 5674

Testing for convergent and divergent series

  1. Given the following geometric series, determine the value of the constant ratio r.

    (2)+4+(8)+16+
    Answer: r=
    numeric
    2 attempts remaining
    STEP: Determine the value of r
    [−2 points ⇒ 0 / 2 points left]

    To calculate the constant ratio (r), we need to determine how consecutive terms in the series are related to each other:

    r=T2T1=42=2Or r=T3T2=84=2

    Therefore, the constant ratio for this geometric series is r=2 .


    Submit your answer as:
  2. Does the given series diverge or converge?

    Answer: The series .
    STEP: Use the test for convergence
    [−1 point ⇒ 0 / 1 points left]

    To test whether an infinte geometric series converges or diverges, we look at the value of the constant ratio:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=2, we can conclude that the series diverges.


    Submit your answer as:

ID is: 2963 Seed is: 6183

Working with geometric series

Two terms in a geometric series, T2 and T5, have values of 1 and 18 respectively. The sum of the first n terms is Sn=341256.

  1. Find the constant ratio for the terms in the series.

    Answer: The constant ratio is: .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We need to find the constant ratio, r, of the series. We have two terms in the series, and we can find r using them.

    T2 and T5 are not successive terms, so we need to put the formula Tn=arn1 to use. In particular, we know that T2=ar and T5=ar4. Now we can create a ratio of these two expressions in order to find r. Start with the ratio T5T2 and then substitute in the expressions with a and r:

    T5T2=ar4ar(18)1=ar4ar

    Now we can simplify the fractions on both sides of the equation. Then solve for r.

    18=r3183=r3312=r

    The constant ratio is r=12.


    Submit your answer as:
  2. Find the value of n. The first term of the series is T1=2.

    Answer: n=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the sum given together with the formula Sn=a(1rn)1r


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 3 / 4 points left]

    We have the sum of the series, so it makes sense to use the formula Sn=a(1rn)1r in order to find the value of n. To get started, substitute all of the values we know into the formula.

    Sn=a(1rn)1r341256=2(1(12)n)1(12)

    STEP: Solve the equation for n
    [−2 points ⇒ 1 / 4 points left]

    Now we need to rearrange the equation in order to find the value of n.

    341256=2(1(12)n)1(12)341256=2(1(12)n)32(341256)(32)=2(1(12)n)(1023512)(12)=1(12)n10231024=1(12)n102310241=(12)n11024=(12)n11024=(12)n

    STEP: Determine n by inspection
    [−1 point ⇒ 0 / 4 points left]

    At this point we must look for a value of n which agrees with the final line. For example, we can try the value n=9 to see if it agrees:

    (12)n(12)9=1512

    That does not agree with the working above, because we need an answer of 11024, not 1512.

    If we continue to try different values for n, we will find that n=10 works:

    (12)n(12)10=11024

    The correct answer for the number of terms in the series is n=10.


    Submit your answer as:

ID is: 2963 Seed is: 1937

Working with geometric series

Two terms in a geometric series, T2 and T3, have values of 1 and 12 respectively. The sum of the first n terms is Sn=6316.

  1. Determine the constant ratio for the terms in the series.

    Answer: The constant ratio is: .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−1 point ⇒ 0 / 1 points left]

    We need to find the constant ratio, r, of the series. We have two terms in the series, and we can find r using them.

    Since T2 and T3 are consecutive terms, we can find r using the relationship r=TnTn1.

    r=TnTn1=T3T2=(12)1=12

    There we have it: the constant ratio is 12.


    Submit your answer as:
  2. Determine the value of n. The first term of the series is T1=2.

    Answer: n=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the sum given together with the formula Sn=a(1rn)1r


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 3 / 4 points left]

    We have the sum of the series, so it makes sense to use the formula Sn=a(1rn)1r in order to find the value of n. To get started, substitute all of the values we know into the formula.

    Sn=a(1rn)1r6316=2(1(12)n)1(12)

    STEP: Solve the equation for n
    [−2 points ⇒ 1 / 4 points left]

    Now we need to rearrange the equation in order to find the value of n.

    6316=2(1(12)n)1(12)6316=2(1(12)n)12(6316)(12)=2(1(12)n)(6332)(12)=1(12)n6364=1(12)n63641=(12)n164=(12)n164=(12)n

    STEP: Determine n by inspection
    [−1 point ⇒ 0 / 4 points left]

    At this point we must look for a value of n which agrees with the final line. For example, we can try the value n=5 to see if it agrees:

    (12)n(12)5=132

    That does not agree with the working above, because we need an answer of 164, not 132.

    If we continue to try different values for n, we will find that n=6 works:

    (12)n(12)6=164

    The correct answer for the number of terms in the series is n=6.


    Submit your answer as:

ID is: 2963 Seed is: 2699

Working with geometric series

A geometric series has the terms T2=16 and T4=64 and a sum of Sn=344.

  1. Determine the value of r for the terms in the series if the value of r is negative.

    Answer: The value of r is: .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We need to find the constant ratio, r, of the series. We have two terms in the series, and we can find r using them.

    T2 and T4 are not successive terms, so we need to put the formula Tn=arn1 to use. In particular, we know that T2=ar and T4=ar3. Now we can create a ratio of these two expressions in order to find r. Start with the ratio T4T2 and then substitute in the expressions with a and r:

    T4T2=ar3ar6416=ar3ar

    Now we can simplify the fractions on both sides of the equation. Then solve for r.

    4=r2±4=r2±2=r

    We get two answers from the calculation. However, remember that the question told us that the constant ratio is negative. Therefore, we should throw out the positve answer. The correct answer for the constant ratio is r=2.


    Submit your answer as:
  2. Determine the number of terms in the sum, n. The first term of the series is T1=8.

    Answer: n=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the sum given together with the formula Sn=a(1rn)1r


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 3 / 4 points left]

    We have the sum of the series, so it makes sense to use the formula Sn=a(1rn)1r in order to find the value of n. To get started, substitute all of the values we know into the formula.

    Sn=a(1rn)1r344=8(1(2)n)1(2)

    STEP: Solve the equation for n
    [−2 points ⇒ 1 / 4 points left]

    Now we need to rearrange the equation in order to find the value of n.

    344=8(1(2)n)1(2)344=8(1(2)n)3(344)(3)=8(1(2)n)(1032)(18)=1(2)n129=1(2)n1291=(2)n128=(2)n128=(2)n

    STEP: Determine n by inspection
    [−1 point ⇒ 0 / 4 points left]

    At this point we must look for a value of n which agrees with the final line. For example, we can try the value n=6 to see if it agrees:

    (2)n(2)6=64

    That does not agree with the working above, because we need an answer of 128, not 64.

    If we continue to try different values for n, we will find that n=7 works:

    (2)n(2)7=128

    The correct answer for the number of terms in the series is n=7.


    Submit your answer as:

ID is: 2961 Seed is: 9511

Using the general formula for the sum of a finite geometric series

A geometric series consists of 4 terms and has a constant ratio of 4. If the first term is 14, find the sum of the series.

Answer: The sum of the series =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Notice that the geometric series consists of a specific number of terms. This means that you have a finite geometric series. Is there a general formula you can use for the sum of a finite geometric series? Do you have the needed values to use this formula?
STEP: Select the version of the general formula to use
[−1 point ⇒ 2 / 3 points left]

We are told in the question that there are a specific number of terms, 4. This means that we are dealing with a finite geometric series. Furthermore, we are given the first term and the constant ratio. We now have enough information to expand the series fully. The first three terms of our series are:

14;1;4;

However, writing out all the terms for a series can be tedious. Luckily, mathematicians have derived a general formula for calculating the sum of a finite geometric series. There are two forms of the general formula:

Sn=a(1rn)1r OR Sn=a(rn1)r1

The first version of the formula is easier to use when r<1 and the second when r>1.

As r<1, we will use the first form of the formula.


STEP: Substitute in the given values
[−2 points ⇒ 0 / 3 points left]

Our final step is to substitute the values given in the question into the appropriate general formula:

Sn=a(1rn)1rS4=14(1(4)4)1(4)S4=514

Therefore, the sum of the geometric series is 514.


Submit your answer as:

ID is: 2961 Seed is: 7884

Using the general formula for the sum of a finite geometric series

Determine the sum of a geometric series given that the first term is 14, the constant ratio is 4 and it consists of 3 terms.

Answer: The sum of the series =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Notice that the geometric series consists of a specific number of terms. This means that you have a finite geometric series. Is there a general formula you can use for the sum of a finite geometric series? Do you have the needed values to use this formula?
STEP: Select the version of the general formula to use
[−1 point ⇒ 2 / 3 points left]

We are told in the question that there are a specific number of terms, 3. This means that we are dealing with a finite geometric series. Furthermore, we are given the first term and the constant ratio. We now have enough information to expand the series fully. The first three terms of our series are:

14;1;4;

However, writing out all the terms for a series can be tedious. Luckily, mathematicians have derived a general formula for calculating the sum of a finite geometric series. There are two forms of the general formula:

Sn=a(1rn)1r OR Sn=a(rn1)r1

The first version of the formula is easier to use when r<1 and the second when r>1.

As r<1, we will use the first form of the formula.


STEP: Substitute in the given values
[−2 points ⇒ 0 / 3 points left]

Our final step is to substitute the values given in the question into the appropriate general formula:

Sn=a(1rn)1rS3=14(1(4)3)1(4)S3=134

Therefore, the sum of the geometric series is 134.


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ID is: 2961 Seed is: 9002

Using the general formula for the sum of a finite geometric series

You are given a geometric series made up of 7 terms with a constant ratio of 2 and the first term equal to 12. Find its sum.

Answer: The sum of the series =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Notice that the geometric series consists of a specific number of terms. This means that you have a finite geometric series. Is there a general formula you can use for the sum of a finite geometric series? Do you have the needed values to use this formula?
STEP: Select the version of the general formula to use
[−1 point ⇒ 2 / 3 points left]

We are told in the question that there are a specific number of terms, 7. This means that we are dealing with a finite geometric series. Furthermore, we are given the first term and the constant ratio. We now have enough information to expand the series fully. The first three terms of our series are:

12;1;2;

However, writing out all the terms for a series can be tedious. Luckily, mathematicians have derived a general formula for calculating the sum of a finite geometric series. There are two forms of the general formula:

Sn=a(1rn)1r OR Sn=a(rn1)r1

The first version of the formula is easier to use when r<1 and the second when r>1.

As r<1, we will use the first form of the formula.


STEP: Substitute in the given values
[−2 points ⇒ 0 / 3 points left]

Our final step is to substitute the values given in the question into the appropriate general formula:

Sn=a(1rn)1rS7=12(1(2)7)1(2)S7=432

Therefore, the sum of the geometric series is 432.


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ID is: 2944 Seed is: 1750

Word problem: determine the maximum height

  1. A plant is 88 cm high and receives an average rainfall of 551 mm. After one year, the plant is 111 cm tall. For the years that follow, the growth of the plant is a third of the previous year's growth.

    Determine the maximum height to which the plant will grow. Give your answer rounded to 1 decimal place.

    Answer:

    Maximum height = cm

    numeric
    2 attempts remaining
    STEP: Write down a series for the annual growth of the plant
    [−2 points ⇒ 2 / 4 points left]

    From the question statement, we know that the plant grows by 111 cm88 cm=23 cm in the first year. Therefore a=23. Be careful not to make the height (111 cm) the first term in the series. It is important to recognise that the series must describe the growth of the plant. We also need to calculate the constant ratio.

    23+233+239+2327+
    r=T2T1=23323=13Or r=T3T2=239233=13

    So we have that a=23 and r=13. This was actually given to us in the question statement:

    third =13=r


    STEP: Apply the condition for convergence to determine the maximum height
    [−2 points ⇒ 0 / 4 points left]

    We notice that with each passing year, the plant grows less and less. We also see that the constant ratio for this series lies within the interval (1;1), which means that the series will converge. In other words, the growth of the plant has a limit or maximum.

    We determine the maximum growth of the plant by calculating the value to which the series converges. To do that, we use the formula for the sum to infinity:

    S=a1r(r1)=23113=2323=34.5

    Therefore, the growth of the plant is limited to 34.5 cm, and the maximum height of the plant is 88 cm+34.5 cm=122.5 cm.


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  2. Determine the general term for the series.

    Answer:

    Tn=

    expression
    2 attempts remaining
    STEP: Write the general formula for the geometric series
    [−1 point ⇒ 0 / 1 points left]
    a=23r=13
    Tn=arn1=23(13)n1

    Therefore, the general term for the series is given by Tn=23(13)n1.

    Although it was not asked for in the question statement, a diagram is shown below to provide a visual representation of this series. It is important to note that we may join the points on the graph because the growth of the plant is continuous.


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ID is: 2944 Seed is: 3839

Word problem: determine the maximum height

  1. A bush is 94 cm high and receives an average rainfall of 718 mm. After one year, the bush is 109 cm tall. For the years that follow, the growth of the bush is a third of the previous year's growth.

    Determine the maximum height to which the bush will grow. Give your answer rounded to 1 decimal place.

    Answer:

    Maximum height = cm

    numeric
    2 attempts remaining
    STEP: Write down a series for the annual growth of the bush
    [−2 points ⇒ 2 / 4 points left]

    From the question statement, we know that the bush grows by 109 cm94 cm=15 cm in the first year. Therefore a=15. Be careful not to make the height (109 cm) the first term in the series. It is important to recognise that the series must describe the growth of the bush. We also need to calculate the constant ratio.

    15+5+53+59+
    r=T2T1=515=13Or r=T3T2=535=13

    So we have that a=15 and r=13. This was actually given to us in the question statement:

    third =13=r


    STEP: Apply the condition for convergence to determine the maximum height
    [−2 points ⇒ 0 / 4 points left]

    We notice that with each passing year, the bush grows less and less. We also see that the constant ratio for this series lies within the interval (1;1), which means that the series will converge. In other words, the growth of the bush has a limit or maximum.

    We determine the maximum growth of the bush by calculating the value to which the series converges. To do that, we use the formula for the sum to infinity:

    S=a1r(r1)=15113=1523=22.5

    Therefore, the growth of the bush is limited to 22.5 cm, and the maximum height of the bush is 94 cm+22.5 cm=116.5 cm.


    Submit your answer as:
  2. Write down the general term for the series.

    Answer:

    Tn=

    expression
    2 attempts remaining
    STEP: Write the general formula for the geometric series
    [−1 point ⇒ 0 / 1 points left]
    a=15r=13
    Tn=arn1=15(13)n1

    Therefore, the general term for the series is given by Tn=15(13)n1.

    Although it was not asked for in the question statement, a diagram is shown below to provide a visual representation of this series. It is important to note that we may join the points on the graph because the growth of the bush is continuous.


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ID is: 2944 Seed is: 3516

Word problem: determine the maximum height

  1. A hedge is 68 cm high and receives an average rainfall of 624 mm. After one year, the hedge is 87 cm tall. For the years that follow, the growth of the hedge is a third of the previous year's growth.

    Determine the maximum height to which the hedge will grow. Give your answer rounded to 2 decimal places.

    Answer:

    Maximum height = cm

    numeric
    2 attempts remaining
    STEP: Write down a series for the annual growth of the hedge
    [−2 points ⇒ 2 / 4 points left]

    From the question statement, we know that the hedge grows by 87 cm68 cm=19 cm in the first year. Therefore a=19. Be careful not to make the height (87 cm) the first term in the series. It is important to recognise that the series must describe the growth of the hedge. We also need to calculate the constant ratio.

    19+193+199+1927+
    r=T2T1=19319=13Or r=T3T2=199193=13

    So we have that a=19 and r=13. This was actually given to us in the question statement:

    third =13=r


    STEP: Apply the condition for convergence to determine the maximum height
    [−2 points ⇒ 0 / 4 points left]

    We notice that with each passing year, the hedge grows less and less. We also see that the constant ratio for this series lies within the interval (1;1), which means that the series will converge. In other words, the growth of the hedge has a limit or maximum.

    We determine the maximum growth of the hedge by calculating the value to which the series converges. To do that, we use the formula for the sum to infinity:

    S=a1r(r1)=19113=1923=28.5

    Therefore, the growth of the hedge is limited to 28.5 cm, and the maximum height of the hedge is 68 cm+28.5 cm=96.5 cm.


    Submit your answer as:
  2. Give the general term for the series.

    Answer:

    Tn=

    expression
    2 attempts remaining
    STEP: Write the general formula for the geometric series
    [−1 point ⇒ 0 / 1 points left]
    a=19r=13
    Tn=arn1=19(13)n1

    Therefore, the general term for the series is given by Tn=19(13)n1.

    Although it was not asked for in the question statement, a diagram is shown below to provide a visual representation of this series. It is important to note that we may join the points on the graph because the growth of the hedge is continuous.


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ID is: 2945 Seed is: 3081

Convergent series: determine the values of k

Calculate for which values of k the geometric series Tn=65n1(k+1)n1 will converge.
Give your answer as an inequality, for example: -1 < x < 1.

Answer:

Range of k:

relational
2 attempts remaining
STEP: Determine the values of a and r
[−2 points ⇒ 2 / 4 points left]

Let's look at the first few terms in the series:

For n=1:T1=6511(k+1)11=6For n=2:T2=6521(k+1)21=65(k+1)For n=3:T3=6531(k+1)31=652(k+1)2=625(k+1)2For n=4:T4=6541(k+1)41=653(k+1)3=6125(k+1)3

From the given series we can see that a=T1=6.

To determine the value of r, we need to determine the ratio between any two consecutive terms in the series:

r=T2T1=T3T2=TnTn1

r=T2T1=65(k+1)6=15(k+1)Or r=T3T2=625(k+1)265(k+1)=15(k+1)

So we have that a=6 and r=15(k+1).


STEP: Apply the condition for convergence to determine the possible values of k
[−2 points ⇒ 0 / 4 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of k using:

1<r<11<15(k+1)<1Multiply through by 5:5<k+1<5Subtract 1:6<k<4

Therefore, for the series to converge, we have determined that 6<k<4.


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ID is: 2945 Seed is: 9775

Convergent series: determine the values of t

Determine for which values of t the geometric series Tn=43n1(t1)n1 will be convergent.
Give your answer as an inequality, for example: -1 < x < 1.

Answer:

Range of t:

relational
2 attempts remaining
STEP: Determine the values of a and r
[−2 points ⇒ 2 / 4 points left]

Let's look at the first few terms in the series:

For n=1:T1=4311(t1)11=4For n=2:T2=4321(t1)21=43(t1)For n=3:T3=4331(t1)31=432(t1)2=49(t1)2For n=4:T4=4341(t1)41=433(t1)3=427(t1)3

From the given series we can see that a=T1=4.

To determine the value of r, we need to determine the ratio between any two consecutive terms in the series:

r=T2T1=T3T2=TnTn1

r=T2T1=43(t1)4=13(t1)Or r=T3T2=49(t1)243(t1)=13(t1)

So we have that a=4 and r=13(t1).


STEP: Apply the condition for convergence to determine the possible values of t
[−2 points ⇒ 0 / 4 points left]

For a geometric series to converge, the value of r must lie within the interval (1;1). We use this fact to determine the possible values of t:

1<r<11<13(t1)<1Multiply through by 3:3<t1<3Add 1:2<t<4

Therefore, for the series to converge, we have determined that 2<t<4.


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ID is: 2945 Seed is: 9089

Convergent series: determine the values of x

Determine for which values of x the geometric series Tn=24n1(x+1)n1 will be convergent.
Give your answer as an inequality, for example: -1 < x < 1.

Answer:

Range of x:

relational
2 attempts remaining
STEP: Determine the values of a and r
[−2 points ⇒ 2 / 4 points left]

Let's look at the first few terms in the series:

For n=1:T1=2411(x+1)11=2For n=2:T2=2421(x+1)21=24(x+1)=12(x+1)For n=3:T3=2431(x+1)31=242(x+1)2=18(x+1)2For n=4:T4=2441(x+1)41=243(x+1)3=132(x+1)3

From the given series we can see that a=T1=2.

To determine the value of r, we need to determine the ratio between any two consecutive terms in the series:

r=T2T1=T3T2=TnTn1

r=T2T1=12(x+1)2=14(x+1)Or r=T3T2=18(x+1)212(x+1)=14(x+1)

So we have that a=2 and r=14(x+1).


STEP: Apply the condition for convergence to determine the possible values of x
[−2 points ⇒ 0 / 4 points left]

For a geometric series to converge, the value of r must lie within the interval (1;1). We use this fact to determine the possible values of x:

1<r<11<14(x+1)<1Multiply through by 4:4<x+1<4Subtract 1:5<x<3

Therefore, for the series to converge, we have determined that 5<x<3.


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ID is: 2956 Seed is: 4347

Finding the r and the Sn for a geometric series

The sixth term of a geometric series is 116. The ninth term is 1128. Answer the questions below about the series.

  1. Determine the constant ratio for the terms in the series.

    Answer: The constant ratio is:
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−3 points ⇒ 0 / 3 points left]

    We have the values for two terms in a geometric series, and we need to find the constant ratio, r, of the series. To begin, it will be helpful to write down what we know: the sixth term is 116, which means T6=116. Similarly, the ninth term is 1128, and we can write T9=1128.

    T6 and T9 are not consecutive terms, so we cannot use the usual formula r=TnTn1 to find r. Instead, we will use the formula Tn=arn1. Based on this formula we can write T6=ar5 and T9=ar8. Now we can create a ratio of these two expressions in order to find r. Start with the ratio T9T6 and then substitute in the expressions with a and r:

    T9T6=ar8ar5(1128)(116)=ar8ar5

    Now we can simplify the fractions on both sides of the equation. Then solve for r.

    18=r3183=r3312=r

    The constant ratio is r=12.


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  2. Determine the sum of the first 5 terms. The first term of the series is T1=2.

    Answer:

    S5=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute a, r and n into the sum formula Sn=a(1rn)1r.


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 2 / 3 points left]

    We can use the formula Sn=a(1rn)1r to find the sum of 5 terms. Substitute all the values we know into the formula.

    Sn=a(1rn)1rS5=2(1(12)5)1(12)

    STEP: Evaluate the formula to get the sum
    [−2 points ⇒ 0 / 3 points left]

    Evaluate the right hand side of the formula. Start inside of the brackets in the numerator and also by evaluating the denominator.

    S5=2(1(12)5)1(12)=2(1(132))32=2(3332)32=(3316)(23)=118

    The sum of the first 5 terms of the series is 118.


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ID is: 2956 Seed is: 9955

Finding the r and the Sn for a geometric series

The second and third terms of a geometric series are 14 and 18 respectively. Answer the questions below about the series.

  1. Compute the constant ratio for the terms in the series.

    Answer: The constant ratio is:
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We have the values for two terms in a geometric series, and we need to find the constant ratio, r, of the series. To begin, it will be helpful to write down what we know: the second term is 14, which means T2=14. Similarly, the third term is 18, and we can write T3=18.

    T2 and T3 are consecutive terms so we can find r using the relationship r=TnTn1.

    r=TnTn1=T3T2=(18)(14)=12

    There we have the answer: the constant ratio is 12.


    Submit your answer as:
  2. What is the sum of the first 10 terms? The first term of the series is T1=12.

    Answer:

    S10=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute a, r and n into the sum formula Sn=a(1rn)1r.


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 2 / 3 points left]

    We can use the formula Sn=a(1rn)1r to find the sum of 10 terms. Substitute all the values we know into the formula.

    Sn=a(1rn)1rS10=12(1(12)10)1(12)

    STEP: Evaluate the formula to get the sum
    [−2 points ⇒ 0 / 3 points left]

    Evaluate the right hand side of the formula. Start inside of the brackets in the numerator and also by evaluating the denominator.

    S10=12(1(12)10)1(12)=12(1(11024))12=12(10231024)12=(10232048)(2)=10231024

    The sum of the first 10 terms of the series is 10231024.


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ID is: 2956 Seed is: 7322

Finding the r and the Sn for a geometric series

Suppose a geometric series has a second term of 1 and a third term of 14. Answer the questions below about the series.

  1. Compute the value of r for the series.

    Answer: The value of r is:
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We have the values for two terms in a geometric series, and we need to find the constant ratio, r, of the series. To begin, it will be helpful to write down what we know: the second term is 1, which means T2=1. Similarly, the third term is 14, and we can write T3=14.

    T2 and T3 are successive terms so we can find r using the relationship r=TnTn1.

    r=TnTn1=T3T2=(14)1=14

    There we have the answer: the value of r is 14.


    Submit your answer as:
  2. Compute the sum of the first 6 terms. The first term of the series is T1=4.

    Answer:

    S6=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute a, r and n into the sum formula Sn=a(1rn)1r.


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 2 / 3 points left]

    We can use the formula Sn=a(1rn)1r to find the sum of 6 terms. Substitute all the values we know into the formula.

    Sn=a(1rn)1rS6=4(1(14)6)1(14)

    STEP: Evaluate the formula to get the sum
    [−2 points ⇒ 0 / 3 points left]

    Evaluate the right hand side of the formula. Start inside of the brackets in the numerator and also by evaluating the denominator.

    S6=4(1(14)6)1(14)=4(1(14096))54=4(40954096)54=(40951024)(45)=819256

    The sum of the first 6 terms of the series is 819256.


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ID is: 2952 Seed is: 9385

Finding the sum to infinity of a geometric series

  1. Given the series, determine the value of r.

    (2)+4+(8)+16+
    Answer: r=
    numeric
    2 attempts remaining
    STEP: Determine the value of r
    [−1 point ⇒ 0 / 1 points left]

    To calculate r, we need to determine the ratio between any two consecutive terms in the series:

    r=T2T1=42=2Or r=T3T2=84=2

    Therefore, we have that r=2.


    Submit your answer as:
  2. Find the sum to infinity.

    INSTRUCTION: If it does not exist, write "Not exist" in the input box.
    Answer: S=
    one-of
    type(string.nocase)
    2 attempts remaining
    STEP: Use the test for convergence to determine if the sum to infinity exists
    [−2 points ⇒ 0 / 2 points left]

    We need to look at the value of the constant ratio to determine whether an infinte geometric series converges or diverges:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=2, we can conclude that the series diverges and sum to infinity does not exist.


    Submit your answer as:

ID is: 2952 Seed is: 4372

Finding the sum to infinity of a geometric series

  1. Given the series, determine the value of r.

    (4)+(8)+(16)+(32)+
    Answer: r=
    numeric
    2 attempts remaining
    STEP: Determine the value of r
    [−1 point ⇒ 0 / 1 points left]

    To calculate r, we need to determine the ratio between any two consecutive terms in the series:

    r=T2T1=84=2Or r=T3T2=168=2

    Therefore, we have that r=2.


    Submit your answer as:
  2. Determine the sum to infinity.

    INSTRUCTION: If it does not exist, write "Not exist" in the input box.
    Answer: S=
    one-of
    type(string.nocase)
    2 attempts remaining
    STEP: Use the test for convergence to determine if the sum to infinity exists
    [−2 points ⇒ 0 / 2 points left]

    We need to look at the value of the constant ratio to determine whether an infinte geometric series converges or diverges:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=2, we can conclude that the series diverges and sum to infinity does not exist.


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ID is: 2952 Seed is: 8915

Finding the sum to infinity of a geometric series

  1. Given the series, determine the value of r.

    4+(16)+64+(256)+
    Answer: r=
    numeric
    2 attempts remaining
    STEP: Determine the value of r
    [−1 point ⇒ 0 / 1 points left]

    To calculate r, we need to determine the ratio between any two consecutive terms in the series:

    r=T2T1=164=4Or r=T3T2=6416=4

    Therefore, we have that r=4.


    Submit your answer as:
  2. Calculate the sum to infinity.

    INSTRUCTION: If it does not exist, write "Not exist" in the input box.
    Answer: S=
    one-of
    type(string.nocase)
    2 attempts remaining
    STEP: Use the test for convergence to determine if the sum to infinity exists
    [−2 points ⇒ 0 / 2 points left]

    We need to look at the value of the constant ratio to determine whether an infinte geometric series converges or diverges:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=4, we can conclude that the series diverges and sum to infinity does not exist.


    Submit your answer as:

ID is: 2943 Seed is: 9776

Convergent series: determine the values of y

Calculate for which values of y the geometric series Sn=3+35(y+1)+352(y+1)2 will be convergent.

INSTRUCTION: Give your answer as an inequality.
Answer: Range of y: .
relational
2 attempts remaining
STEP: Determine the value of r
[−1 point ⇒ 2 / 3 points left]

A geometric series converges (has a finite sum) if the constant ratio is between 1 and 1. So we need to determine the value of r for this series. As always, the ratio comes from any two consecutive terms:

r=T2T1=T3T2=TnTn1

r=T2T1=35(y+1)3=15(y+1)Or r=T3T2=352(y+1)235(y+1)=15(y+1)

So we know that r=15(y+1).


STEP: Apply the condition for convergence to determine the possible values of y
[−2 points ⇒ 0 / 3 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of y using:

1<r<11<15(y+1)<1Multiply through by 5:5<y+1<5Subtract 1:6<y<4

We have the answer: for the series to converge, y must be in this range: 6<y<4.


Submit your answer as:

ID is: 2943 Seed is: 3300

Convergent series: determine the values of t

Calculate for which values of t the geometric series Sn=2+27(t+1)+272(t+1)2 will be convergent.

INSTRUCTION: Give your answer as an inequality.
Answer: Range of t: .
relational
2 attempts remaining
STEP: Determine the value of r
[−1 point ⇒ 2 / 3 points left]

A geometric series converges (has a finite sum) if the constant ratio is between 1 and 1. So we need to determine the value of r for this series. As always, the ratio comes from any two consecutive terms:

r=T2T1=T3T2=TnTn1

r=T2T1=27(t+1)2=17(t+1)Or r=T3T2=272(t+1)227(t+1)=17(t+1)

So we know that r=17(t+1).


STEP: Apply the condition for convergence to determine the possible values of t
[−2 points ⇒ 0 / 3 points left]

For a geometric series to converge, the value of r must lie within the interval (1;1). We use this fact to determine the possible values of t:

1<r<11<17(t+1)<1Multiply through by 7:7<t+1<7Subtract 1:8<t<6

We have the answer: for the series to converge, t must be in this range: 8<t<6.


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ID is: 2943 Seed is: 1913

Convergent series: determine the values of v

Determine for which values of v the geometric series Sn=6+63(v1)+632(v1)2 will converge.

INSTRUCTION: Give your answer as an inequality.
Answer: Range of v: .
relational
2 attempts remaining
STEP: Determine the value of r
[−1 point ⇒ 2 / 3 points left]

A geometric series converges (has a finite sum) if the constant ratio is between 1 and 1. So we need to determine the value of r for this series. As always, the ratio comes from any two consecutive terms:

r=T2T1=T3T2=TnTn1

r=T2T1=63(v1)6=13(v1)Or r=T3T2=632(v1)263(v1)=13(v1)

So we know that r=13(v1).


STEP: Apply the condition for convergence to determine the possible values of v
[−2 points ⇒ 0 / 3 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of v using:

1<r<11<13(v1)<1Multiply through by 3:3<v1<3Add 1:2<v<4

We have the answer: for the series to converge, v must be in this range: 2<v<4.


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ID is: 3924 Seed is: 4678

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2
Maths formulas

Given the geometric sequence:

18;9;4.5;2.25;
  1. Complete the statement below about the following infinite series:

    18+9+4.5+2.25+
    Answer: The series converge because . 2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise infinite series in the Everything Maths textbook.


    STEP: Test the series against conditions for convergence
    [−2 points ⇒ 0 / 2 points left]

    Whether or not a geometric series will converge depends on the value of the constant ratio r. If 1<r<1, the series will converge and the sum of the terms will tend towards some finite value. Otherwise, the sum of the terms will be infinite.

    We work out the value of r by taking the ratio of any two consecutive terms in the sequence.

    r=T2T1=918=12

    This value of r is 1<12<1.

    So the series will converge because 1<r<1.


    Submit your answer as: and
  2. We can find an expression for SSn in the form pqn, where Sn is the sum of the first n terms of the given geometric sequence. What are the values of p and q?

    INSTRUCTION: Type the values of p and q into the boxes below. Do not include any brackets.
    Answer:
    1. p=
    2. q=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Determine the values of S and Sn, and then simplify to find an expression of the form abn.


    STEP: Determine the values of S and Sn
    [−2 points ⇒ 2 / 4 points left]
    S=a1r;1<r<1

    Substitute in the values a=18 and r=12:

    S=a1r=18112=1812=36

    and for Sn:

    Sn=a(rn1)r1;r1=18((12)n1)121=18((12)n1)12=36((12)n1)

    STEP: Determine the values of p and q
    [−2 points ⇒ 0 / 4 points left]

    We can now subtract Sn from S and simplify:

    SSn=36(36((12)n1))=36+36(12)n36=36(12)n

    This is in the correct form pqn. So the correct answers are:

    1. p=36
    2. q=12

    Submit your answer as: and

ID is: 3924 Seed is: 1246

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2
Maths formulas

Given the geometric sequence:

18;9;4.5;2.25;
  1. Complete the statement below about the following infinite series:

    18+9+4.5+2.25+
    Answer: The series converge because . 2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise infinite series in the Everything Maths textbook.


    STEP: Test the series against conditions for convergence
    [−2 points ⇒ 0 / 2 points left]

    Whether or not a geometric series will converge depends on the value of the constant ratio r. If 1<r<1, the series will converge and the sum of the terms will tend towards some finite value. Otherwise, the sum of the terms will be infinite.

    We work out the value of r by taking the ratio of any two consecutive terms in the sequence.

    r=T2T1=918=12

    This value of r is 1<12<1.

    So the series will converge because 1<r<1.


    Submit your answer as: and
  2. We can find an expression for SSn in the form pqn, where Sn is the sum of the first n terms of the given geometric sequence. What are the values of p and q?

    INSTRUCTION: Type the values of p and q into the boxes below. Do not include any brackets.
    Answer:
    1. p=
    2. q=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Determine the values of S and Sn, and then simplify to find an expression of the form abn.


    STEP: Determine the values of S and Sn
    [−2 points ⇒ 2 / 4 points left]
    S=a1r;1<r<1

    Substitute in the values a=18 and r=12:

    S=a1r=18112=1812=36

    and for Sn:

    Sn=a(rn1)r1;r1=18((12)n1)121=18((12)n1)12=36((12)n1)

    STEP: Determine the values of p and q
    [−2 points ⇒ 0 / 4 points left]

    We can now subtract Sn from S and simplify:

    SSn=36(36((12)n1))=36+36(12)n36=36(12)n

    This is in the correct form pqn. So the correct answers are:

    1. p=36
    2. q=12

    Submit your answer as: and

ID is: 3924 Seed is: 3091

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2
Maths formulas

Given the geometric sequence:

12;6;3;1.5;
  1. Complete the statement below about the following infinite series:

    12+6+3+1.5+
    Answer: The series converge because . 2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise infinite series in the Everything Maths textbook.


    STEP: Test the series against conditions for convergence
    [−2 points ⇒ 0 / 2 points left]

    Whether or not a geometric series will converge depends on the value of the constant ratio r. If 1<r<1, the series will converge and the sum of the terms will tend towards some finite value. Otherwise, the sum of the terms will be infinite.

    We work out the value of r by taking the ratio of any two consecutive terms in the sequence.

    r=T2T1=612=12

    This value of r is 1<12<1.

    So the series will converge because 1<r<1.


    Submit your answer as: and
  2. We can find an expression for SSn in the form pqn, where Sn is the sum of the first n terms of the given geometric sequence. What are the values of p and q?

    INSTRUCTION: Type the values of p and q into the boxes below. Do not include any brackets.
    Answer:
    1. p=
    2. q=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Determine the values of S and Sn, and then simplify to find an expression of the form abn.


    STEP: Determine the values of S and Sn
    [−2 points ⇒ 2 / 4 points left]
    S=a1r;1<r<1

    Substitute in the values a=12 and r=12:

    S=a1r=12112=1212=24

    and for Sn:

    Sn=a(rn1)r1;r1=12((12)n1)121=12((12)n1)12=24((12)n1)

    STEP: Determine the values of p and q
    [−2 points ⇒ 0 / 4 points left]

    We can now subtract Sn from S and simplify:

    SSn=24(24((12)n1))=24+24(12)n24=24(12)n

    This is in the correct form pqn. So the correct answers are:

    1. p=24
    2. q=12

    Submit your answer as: and

ID is: 2936 Seed is: 7953

Convergent and divergent series

  1. Given the general term Tn=(67)n, determine an expression for the sum of the first n terms of the series.

    Answer: Sn=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    What kind of series does this general term give us?


    STEP: Use the general term to write down the first few terms in the series
    [−2 points ⇒ 1 / 3 points left]

    Consider the general term Tn=(67)n. We need to determine if this will generate an arithmetic or geometric series, so we write down the first few terms in the series:

    For n=1:T1=(67)1=67For n=2:T2=(67)2=3649For n=3:T3=(67)3=216343For n=4:T4=(67)4=12962401

    Therefore, we have the following series:

    67+3649+216343+12962401+

    So, the first term for the series is a=67.

    Now we need to see if there is a constant ratio (r), which would make this a geometric series:

    r=T2T1=364967=67Or r=T3T2=2163433649=67

    Therefore, this is a geometric series with a=67 and r=67.


    STEP: Determine an expression for the sum of the first n terms of the series
    [−1 point ⇒ 0 / 3 points left]

    Since we know that we are dealing with a geometric series and that 0<r<1, we use the following formula:

    Sn=a(1rn)1r

    We substitute in the known values and simplify:

    Sn=67(1(67)n)167=67(1(67)n)17=6717(1(67)n)=6(1(67)n)

    Therefore, the sum of the first n terms of the series is given by 6(1(67)n).


    Submit your answer as:
  2. Is the series convergent or divergent? Use the expression for the sum of the first n terms of the series to complete the following sentence:

    Answer:

    We can conclude that the given series is because:

    2 attempts remaining
    STEP: Consider the sum to infinity
    [−2 points ⇒ 0 / 2 points left]

    To complete the sentence given in the question statement, we look at the expression Sn=6(1(67)n) more closely and consider what happens to the sum as n tends to infinity.

    Let's break the expression down a bit:

    As n:(67)n01(67)n16(1(67)n)6Sn6

    Therefore, we know that the series is convergent because Sn6 as n.


    Submit your answer as: and

ID is: 2936 Seed is: 3023

Convergent and divergent series

  1. Consider the general term Tn=5×4n+1, determine an expression for the sum of the first n terms of the series.

    Answer: Sn=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    What kind of series does this general term give us?


    STEP: Use the general term to write down the first few terms in the series
    [−2 points ⇒ 1 / 3 points left]

    Consider the general term Tn=5×4n+1. We need to determine if this will generate an arithmetic or geometric series, so we write down the first few terms in the series:

    For n=1:T1=5×42=80For n=2:T2=5×43=320For n=3:T3=5×44=1280For n=4:T4=5×45=5120

    Therefore, we have the following series:

    80+320+1280+5120+

    So, the first term for the series is a=80.

    Now we need to see if there is a constant ratio (r), which would make this a geometric series:

    r=T2T1=32080=4Or r=T3T2=1280320=4

    Therefore, this is a geometric series with a=80 and r=4.


    STEP: Determine an expression for the sum of the first n terms of the series
    [−1 point ⇒ 0 / 3 points left]

    Since we know that we are dealing with a geometric series and that r>1, we use the following formula:

    Sn=a(rn1)r1

    We substitute in the known values and simplify:

    Sn=80(4n1)41=80(4n1)3=803(4n1)

    Therefore, the sum of the first n terms of the series is given by 803(4n1).


    Submit your answer as:
  2. Is the series convergent or divergent? Use the expression for the sum of the first n terms of the series to complete the following sentence:

    Answer:

    We can conclude that the given series is because:

    2 attempts remaining
    STEP: Consider the sum to infinity
    [−2 points ⇒ 0 / 2 points left]

    To complete the sentence given in the question statement, we look at the expression Sn=803(4n1) more closely and consider what happens to the sum as n tends to infinity.

    Let's break the expression down a bit:

    As n:4n4n1803(4n1)Sn

    Therefore, we know that the series is divergent because Sn as n.


    Submit your answer as: and

ID is: 2936 Seed is: 6350

Convergent and divergent series

  1. Given the general term Tn=(17)n, determine an expression for the sum of the first n terms of the series.

    Answer: Sn=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    What kind of series does this general term give us?


    STEP: Use the general term to write down the first few terms in the series
    [−2 points ⇒ 1 / 3 points left]

    Consider the general term Tn=(17)n. We need to determine if this will generate an arithmetic or geometric series, so we write down the first few terms in the series:

    For n=1:T1=(17)1=17For n=2:T2=(17)2=149For n=3:T3=(17)3=1343For n=4:T4=(17)4=12401

    Therefore, we have the following series:

    17+149+1343+12401+

    So, the first term for the series is a=17.

    Now we need to see if there is a constant ratio (r), which would make this a geometric series:

    r=T2T1=14917=17Or r=T3T2=1343149=17

    Therefore, this is a geometric series with a=17 and r=17.


    STEP: Determine an expression for the sum of the first n terms of the series
    [−1 point ⇒ 0 / 3 points left]

    Since we know that we are dealing with a geometric series and that 0<r<1, we use the following formula:

    Sn=a(1rn)1r

    We substitute in the known values and simplify:

    Sn=17(1(17)n)117=17(1(17)n)67=1767(1(17)n)=16(1(17)n)

    Therefore, the sum of the first n terms of the series is given by 16(1(17)n).


    Submit your answer as:
  2. Is the series convergent or divergent? Use the expression for the sum of the first n terms of the series to complete the following sentence:

    Answer:

    We can conclude that the given series is because:

    2 attempts remaining
    STEP: Consider the sum to infinity
    [−2 points ⇒ 0 / 2 points left]

    To complete the sentence given in the question statement, we look at the expression Sn=16(1(17)n) more closely and consider what happens to the sum as n tends to infinity.

    Let's break the expression down a bit:

    As n:(17)n01(17)n116(1(17)n)16Sn16

    Therefore, we know that the series is convergent because as n gets larger and larger, Sn tends to 16.


    Submit your answer as: and